Related
I would like to read a resource from within my jar like so:
File file;
file = new File(getClass().getResource("/file.txt").toURI());
BufferedReader reader = new BufferedReader(new FileReader(file));
//Read the file
and it works fine when running it in Eclipse, but if I export it to a jar, and then run it, there is an IllegalArgumentException:
Exception in thread "Thread-2"
java.lang.IllegalArgumentException: URI is not hierarchical
and I really don't know why but with some testing I found if I change
file = new File(getClass().getResource("/file.txt").toURI());
to
file = new File(getClass().getResource("/folder/file.txt").toURI());
then it works the opposite (it works in jar but not eclipse).
I'm using Eclipse and the folder with my file is in a class folder.
Rather than trying to address the resource as a File just ask the ClassLoader to return an InputStream for the resource instead via getResourceAsStream:
try (InputStream in = getClass().getResourceAsStream("/file.txt");
BufferedReader reader = new BufferedReader(new InputStreamReader(in))) {
// Use resource
}
As long as the file.txt resource is available on the classpath then this approach will work the same way regardless of whether the file.txt resource is in a classes/ directory or inside a jar.
The URI is not hierarchical occurs because the URI for a resource within a jar file is going to look something like this: file:/example.jar!/file.txt. You cannot read the entries within a jar (a zip file) like it was a plain old File.
This is explained well by the answers to:
How do I read a resource file from a Java jar file?
Java Jar file: use resource errors: URI is not hierarchical
To access a file in a jar you have two options:
Place the file in directory structure matching your package name (after extracting .jar file, it should be in the same directory as .class file), then access it using getClass().getResourceAsStream("file.txt")
Place the file at the root (after extracting .jar file, it should be in the root), then access it using Thread.currentThread().getContextClassLoader().getResourceAsStream("file.txt")
The first option may not work when jar is used as a plugin.
I had this problem before and I made fallback way for loading. Basically first way work within .jar file and second way works within eclipse or other IDE.
public class MyClass {
public static InputStream accessFile() {
String resource = "my-file-located-in-resources.txt";
// this is the path within the jar file
InputStream input = MyClass.class.getResourceAsStream("/resources/" + resource);
if (input == null) {
// this is how we load file within editor (eg eclipse)
input = MyClass.class.getClassLoader().getResourceAsStream(resource);
}
return input;
}
}
Up until now (December 2017), this is the only solution I found which works both inside and outside the IDE.
Use PathMatchingResourcePatternResolver
Note: it works also in spring-boot
In this example I'm reading some files located in src/main/resources/my_folder:
try {
// Get all the files under this inner resource folder: my_folder
String scannedPackage = "my_folder/*";
PathMatchingResourcePatternResolver scanner = new PathMatchingResourcePatternResolver();
Resource[] resources = scanner.getResources(scannedPackage);
if (resources == null || resources.length == 0)
log.warn("Warning: could not find any resources in this scanned package: " + scannedPackage);
else {
for (Resource resource : resources) {
log.info(resource.getFilename());
// Read the file content (I used BufferedReader, but there are other solutions for that):
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(resource.getInputStream()));
String line = null;
while ((line = bufferedReader.readLine()) != null) {
// ...
// ...
}
bufferedReader.close();
}
}
} catch (Exception e) {
throw new Exception("Failed to read the resources folder: " + e.getMessage(), e);
}
The problem is that certain third party libraries require file pathnames rather than input streams. Most of the answers don't address this issue.
In this case, one workaround is to copy the resource contents into a temporary file. The following example uses jUnit's TemporaryFolder.
private List<String> decomposePath(String path){
List<String> reversed = Lists.newArrayList();
File currFile = new File(path);
while(currFile != null){
reversed.add(currFile.getName());
currFile = currFile.getParentFile();
}
return Lists.reverse(reversed);
}
private String writeResourceToFile(String resourceName) throws IOException {
ClassLoader loader = getClass().getClassLoader();
InputStream configStream = loader.getResourceAsStream(resourceName);
List<String> pathComponents = decomposePath(resourceName);
folder.newFolder(pathComponents.subList(0, pathComponents.size() - 1).toArray(new String[0]));
File tmpFile = folder.newFile(resourceName);
Files.copy(configStream, tmpFile.toPath(), REPLACE_EXISTING);
return tmpFile.getAbsolutePath();
}
In my case I finally made it with
import java.lang.Thread;
import java.io.BufferedReader;
import java.io.InputStreamReader;
final BufferedReader in = new BufferedReader(new InputStreamReader(
Thread.currentThread().getContextClassLoader().getResourceAsStream("file.txt"))
); // no initial slash in file.txt
Make sure that you work with the correct separator. I replaced all / in a relative path with a File.separator. This worked fine in the IDE, however did not work in the build JAR.
I have found a fix
BufferedReader br = new BufferedReader(new InputStreamReader(Main.class.getResourceAsStream(path)));
Replace "Main" with the java class you coded it in. replace "path" with the path within the jar file.
for example, if you put State1.txt in the package com.issac.state, then type the path as "/com/issac/state/State1" if you run Linux or Mac. If you run Windows then type the path as "\com\issac\state\State1". Don't add the .txt extension to the file unless the File not found exception occurs.
This code works both in Eclipse and in Exported Runnable JAR
private String writeResourceToFile(String resourceName) throws IOException {
File outFile = new File(certPath + File.separator + resourceName);
if (outFile.isFile())
return outFile.getAbsolutePath();
InputStream resourceStream = null;
// Java: In caso di JAR dentro il JAR applicativo
URLClassLoader urlClassLoader = (URLClassLoader)Cypher.class.getClassLoader();
URL url = urlClassLoader.findResource(resourceName);
if (url != null) {
URLConnection conn = url.openConnection();
if (conn != null) {
resourceStream = conn.getInputStream();
}
}
if (resourceStream != null) {
Files.copy(resourceStream, outFile.toPath(), StandardCopyOption.REPLACE_EXISTING);
return outFile.getAbsolutePath();
} else {
System.out.println("Embedded Resource " + resourceName + " not found.");
}
return "";
}
finally i solved errors:
String input_path = "resources\\file.txt";
input_path = input_path.replace("\\", "/"); // doesn't work with back slash
URL file_url = getClass().getClassLoader().getResource(input_path);
String file_path = new URI(file_url.toString().replace(" ","%20")).getSchemeSpecificPart();
InputStream file_inputStream = file_url.openStream();
You can use class loader which will read from classpath as ROOT path (without "/" in the beginning)
InputStream in = getClass().getClassLoader().getResourceAsStream("file.txt");
BufferedReader reader = new BufferedReader(new InputStreamReader(in));
For some reason classLoader.getResource() always returned null when I deployed the web application to WildFly 14. getting classLoader from getClass().getClassLoader() or Thread.currentThread().getContextClassLoader() returns null.
getClass().getClassLoader() API doc says,
"Returns the class loader for the class. Some implementations may use null to represent the bootstrap class loader. This method will return null in such implementations if this class was loaded by the bootstrap class loader."
may be if you are using WildFly and yours web application try this
request.getServletContext().getResource() returned the resource url. Here request is an object of ServletRequest.
If you are using spring, then you can use the the following method to read file from src/main/resources:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import org.springframework.core.io.ClassPathResource;
public String readFileToString(String path) throws IOException {
StringBuilder resultBuilder = new StringBuilder("");
ClassPathResource resource = new ClassPathResource(path);
try (
InputStream inputStream = resource.getInputStream();
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream))) {
String line;
while ((line = bufferedReader.readLine()) != null) {
resultBuilder.append(line);
}
}
return resultBuilder.toString();
}
Below code works with Spring boot(kotlin):
val authReader = InputStreamReader(javaClass.getResourceAsStream("/file1.json"))
If you wanna read as a file, I believe there still is a similar solution:
ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("file/test.xml").getFile());
I would like to read a resource from within my jar like so:
File file;
file = new File(getClass().getResource("/file.txt").toURI());
BufferedReader reader = new BufferedReader(new FileReader(file));
//Read the file
and it works fine when running it in Eclipse, but if I export it to a jar, and then run it, there is an IllegalArgumentException:
Exception in thread "Thread-2"
java.lang.IllegalArgumentException: URI is not hierarchical
and I really don't know why but with some testing I found if I change
file = new File(getClass().getResource("/file.txt").toURI());
to
file = new File(getClass().getResource("/folder/file.txt").toURI());
then it works the opposite (it works in jar but not eclipse).
I'm using Eclipse and the folder with my file is in a class folder.
Rather than trying to address the resource as a File just ask the ClassLoader to return an InputStream for the resource instead via getResourceAsStream:
try (InputStream in = getClass().getResourceAsStream("/file.txt");
BufferedReader reader = new BufferedReader(new InputStreamReader(in))) {
// Use resource
}
As long as the file.txt resource is available on the classpath then this approach will work the same way regardless of whether the file.txt resource is in a classes/ directory or inside a jar.
The URI is not hierarchical occurs because the URI for a resource within a jar file is going to look something like this: file:/example.jar!/file.txt. You cannot read the entries within a jar (a zip file) like it was a plain old File.
This is explained well by the answers to:
How do I read a resource file from a Java jar file?
Java Jar file: use resource errors: URI is not hierarchical
To access a file in a jar you have two options:
Place the file in directory structure matching your package name (after extracting .jar file, it should be in the same directory as .class file), then access it using getClass().getResourceAsStream("file.txt")
Place the file at the root (after extracting .jar file, it should be in the root), then access it using Thread.currentThread().getContextClassLoader().getResourceAsStream("file.txt")
The first option may not work when jar is used as a plugin.
I had this problem before and I made fallback way for loading. Basically first way work within .jar file and second way works within eclipse or other IDE.
public class MyClass {
public static InputStream accessFile() {
String resource = "my-file-located-in-resources.txt";
// this is the path within the jar file
InputStream input = MyClass.class.getResourceAsStream("/resources/" + resource);
if (input == null) {
// this is how we load file within editor (eg eclipse)
input = MyClass.class.getClassLoader().getResourceAsStream(resource);
}
return input;
}
}
Up until now (December 2017), this is the only solution I found which works both inside and outside the IDE.
Use PathMatchingResourcePatternResolver
Note: it works also in spring-boot
In this example I'm reading some files located in src/main/resources/my_folder:
try {
// Get all the files under this inner resource folder: my_folder
String scannedPackage = "my_folder/*";
PathMatchingResourcePatternResolver scanner = new PathMatchingResourcePatternResolver();
Resource[] resources = scanner.getResources(scannedPackage);
if (resources == null || resources.length == 0)
log.warn("Warning: could not find any resources in this scanned package: " + scannedPackage);
else {
for (Resource resource : resources) {
log.info(resource.getFilename());
// Read the file content (I used BufferedReader, but there are other solutions for that):
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(resource.getInputStream()));
String line = null;
while ((line = bufferedReader.readLine()) != null) {
// ...
// ...
}
bufferedReader.close();
}
}
} catch (Exception e) {
throw new Exception("Failed to read the resources folder: " + e.getMessage(), e);
}
The problem is that certain third party libraries require file pathnames rather than input streams. Most of the answers don't address this issue.
In this case, one workaround is to copy the resource contents into a temporary file. The following example uses jUnit's TemporaryFolder.
private List<String> decomposePath(String path){
List<String> reversed = Lists.newArrayList();
File currFile = new File(path);
while(currFile != null){
reversed.add(currFile.getName());
currFile = currFile.getParentFile();
}
return Lists.reverse(reversed);
}
private String writeResourceToFile(String resourceName) throws IOException {
ClassLoader loader = getClass().getClassLoader();
InputStream configStream = loader.getResourceAsStream(resourceName);
List<String> pathComponents = decomposePath(resourceName);
folder.newFolder(pathComponents.subList(0, pathComponents.size() - 1).toArray(new String[0]));
File tmpFile = folder.newFile(resourceName);
Files.copy(configStream, tmpFile.toPath(), REPLACE_EXISTING);
return tmpFile.getAbsolutePath();
}
In my case I finally made it with
import java.lang.Thread;
import java.io.BufferedReader;
import java.io.InputStreamReader;
final BufferedReader in = new BufferedReader(new InputStreamReader(
Thread.currentThread().getContextClassLoader().getResourceAsStream("file.txt"))
); // no initial slash in file.txt
Make sure that you work with the correct separator. I replaced all / in a relative path with a File.separator. This worked fine in the IDE, however did not work in the build JAR.
I have found a fix
BufferedReader br = new BufferedReader(new InputStreamReader(Main.class.getResourceAsStream(path)));
Replace "Main" with the java class you coded it in. replace "path" with the path within the jar file.
for example, if you put State1.txt in the package com.issac.state, then type the path as "/com/issac/state/State1" if you run Linux or Mac. If you run Windows then type the path as "\com\issac\state\State1". Don't add the .txt extension to the file unless the File not found exception occurs.
This code works both in Eclipse and in Exported Runnable JAR
private String writeResourceToFile(String resourceName) throws IOException {
File outFile = new File(certPath + File.separator + resourceName);
if (outFile.isFile())
return outFile.getAbsolutePath();
InputStream resourceStream = null;
// Java: In caso di JAR dentro il JAR applicativo
URLClassLoader urlClassLoader = (URLClassLoader)Cypher.class.getClassLoader();
URL url = urlClassLoader.findResource(resourceName);
if (url != null) {
URLConnection conn = url.openConnection();
if (conn != null) {
resourceStream = conn.getInputStream();
}
}
if (resourceStream != null) {
Files.copy(resourceStream, outFile.toPath(), StandardCopyOption.REPLACE_EXISTING);
return outFile.getAbsolutePath();
} else {
System.out.println("Embedded Resource " + resourceName + " not found.");
}
return "";
}
finally i solved errors:
String input_path = "resources\\file.txt";
input_path = input_path.replace("\\", "/"); // doesn't work with back slash
URL file_url = getClass().getClassLoader().getResource(input_path);
String file_path = new URI(file_url.toString().replace(" ","%20")).getSchemeSpecificPart();
InputStream file_inputStream = file_url.openStream();
You can use class loader which will read from classpath as ROOT path (without "/" in the beginning)
InputStream in = getClass().getClassLoader().getResourceAsStream("file.txt");
BufferedReader reader = new BufferedReader(new InputStreamReader(in));
For some reason classLoader.getResource() always returned null when I deployed the web application to WildFly 14. getting classLoader from getClass().getClassLoader() or Thread.currentThread().getContextClassLoader() returns null.
getClass().getClassLoader() API doc says,
"Returns the class loader for the class. Some implementations may use null to represent the bootstrap class loader. This method will return null in such implementations if this class was loaded by the bootstrap class loader."
may be if you are using WildFly and yours web application try this
request.getServletContext().getResource() returned the resource url. Here request is an object of ServletRequest.
If you are using spring, then you can use the the following method to read file from src/main/resources:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import org.springframework.core.io.ClassPathResource;
public String readFileToString(String path) throws IOException {
StringBuilder resultBuilder = new StringBuilder("");
ClassPathResource resource = new ClassPathResource(path);
try (
InputStream inputStream = resource.getInputStream();
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream))) {
String line;
while ((line = bufferedReader.readLine()) != null) {
resultBuilder.append(line);
}
}
return resultBuilder.toString();
}
Below code works with Spring boot(kotlin):
val authReader = InputStreamReader(javaClass.getResourceAsStream("/file1.json"))
If you wanna read as a file, I believe there still is a similar solution:
ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("file/test.xml").getFile());
I would like to read a resource from within my jar like so:
File file;
file = new File(getClass().getResource("/file.txt").toURI());
BufferedReader reader = new BufferedReader(new FileReader(file));
//Read the file
and it works fine when running it in Eclipse, but if I export it to a jar, and then run it, there is an IllegalArgumentException:
Exception in thread "Thread-2"
java.lang.IllegalArgumentException: URI is not hierarchical
and I really don't know why but with some testing I found if I change
file = new File(getClass().getResource("/file.txt").toURI());
to
file = new File(getClass().getResource("/folder/file.txt").toURI());
then it works the opposite (it works in jar but not eclipse).
I'm using Eclipse and the folder with my file is in a class folder.
Rather than trying to address the resource as a File just ask the ClassLoader to return an InputStream for the resource instead via getResourceAsStream:
try (InputStream in = getClass().getResourceAsStream("/file.txt");
BufferedReader reader = new BufferedReader(new InputStreamReader(in))) {
// Use resource
}
As long as the file.txt resource is available on the classpath then this approach will work the same way regardless of whether the file.txt resource is in a classes/ directory or inside a jar.
The URI is not hierarchical occurs because the URI for a resource within a jar file is going to look something like this: file:/example.jar!/file.txt. You cannot read the entries within a jar (a zip file) like it was a plain old File.
This is explained well by the answers to:
How do I read a resource file from a Java jar file?
Java Jar file: use resource errors: URI is not hierarchical
To access a file in a jar you have two options:
Place the file in directory structure matching your package name (after extracting .jar file, it should be in the same directory as .class file), then access it using getClass().getResourceAsStream("file.txt")
Place the file at the root (after extracting .jar file, it should be in the root), then access it using Thread.currentThread().getContextClassLoader().getResourceAsStream("file.txt")
The first option may not work when jar is used as a plugin.
I had this problem before and I made fallback way for loading. Basically first way work within .jar file and second way works within eclipse or other IDE.
public class MyClass {
public static InputStream accessFile() {
String resource = "my-file-located-in-resources.txt";
// this is the path within the jar file
InputStream input = MyClass.class.getResourceAsStream("/resources/" + resource);
if (input == null) {
// this is how we load file within editor (eg eclipse)
input = MyClass.class.getClassLoader().getResourceAsStream(resource);
}
return input;
}
}
Up until now (December 2017), this is the only solution I found which works both inside and outside the IDE.
Use PathMatchingResourcePatternResolver
Note: it works also in spring-boot
In this example I'm reading some files located in src/main/resources/my_folder:
try {
// Get all the files under this inner resource folder: my_folder
String scannedPackage = "my_folder/*";
PathMatchingResourcePatternResolver scanner = new PathMatchingResourcePatternResolver();
Resource[] resources = scanner.getResources(scannedPackage);
if (resources == null || resources.length == 0)
log.warn("Warning: could not find any resources in this scanned package: " + scannedPackage);
else {
for (Resource resource : resources) {
log.info(resource.getFilename());
// Read the file content (I used BufferedReader, but there are other solutions for that):
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(resource.getInputStream()));
String line = null;
while ((line = bufferedReader.readLine()) != null) {
// ...
// ...
}
bufferedReader.close();
}
}
} catch (Exception e) {
throw new Exception("Failed to read the resources folder: " + e.getMessage(), e);
}
The problem is that certain third party libraries require file pathnames rather than input streams. Most of the answers don't address this issue.
In this case, one workaround is to copy the resource contents into a temporary file. The following example uses jUnit's TemporaryFolder.
private List<String> decomposePath(String path){
List<String> reversed = Lists.newArrayList();
File currFile = new File(path);
while(currFile != null){
reversed.add(currFile.getName());
currFile = currFile.getParentFile();
}
return Lists.reverse(reversed);
}
private String writeResourceToFile(String resourceName) throws IOException {
ClassLoader loader = getClass().getClassLoader();
InputStream configStream = loader.getResourceAsStream(resourceName);
List<String> pathComponents = decomposePath(resourceName);
folder.newFolder(pathComponents.subList(0, pathComponents.size() - 1).toArray(new String[0]));
File tmpFile = folder.newFile(resourceName);
Files.copy(configStream, tmpFile.toPath(), REPLACE_EXISTING);
return tmpFile.getAbsolutePath();
}
In my case I finally made it with
import java.lang.Thread;
import java.io.BufferedReader;
import java.io.InputStreamReader;
final BufferedReader in = new BufferedReader(new InputStreamReader(
Thread.currentThread().getContextClassLoader().getResourceAsStream("file.txt"))
); // no initial slash in file.txt
Make sure that you work with the correct separator. I replaced all / in a relative path with a File.separator. This worked fine in the IDE, however did not work in the build JAR.
I have found a fix
BufferedReader br = new BufferedReader(new InputStreamReader(Main.class.getResourceAsStream(path)));
Replace "Main" with the java class you coded it in. replace "path" with the path within the jar file.
for example, if you put State1.txt in the package com.issac.state, then type the path as "/com/issac/state/State1" if you run Linux or Mac. If you run Windows then type the path as "\com\issac\state\State1". Don't add the .txt extension to the file unless the File not found exception occurs.
This code works both in Eclipse and in Exported Runnable JAR
private String writeResourceToFile(String resourceName) throws IOException {
File outFile = new File(certPath + File.separator + resourceName);
if (outFile.isFile())
return outFile.getAbsolutePath();
InputStream resourceStream = null;
// Java: In caso di JAR dentro il JAR applicativo
URLClassLoader urlClassLoader = (URLClassLoader)Cypher.class.getClassLoader();
URL url = urlClassLoader.findResource(resourceName);
if (url != null) {
URLConnection conn = url.openConnection();
if (conn != null) {
resourceStream = conn.getInputStream();
}
}
if (resourceStream != null) {
Files.copy(resourceStream, outFile.toPath(), StandardCopyOption.REPLACE_EXISTING);
return outFile.getAbsolutePath();
} else {
System.out.println("Embedded Resource " + resourceName + " not found.");
}
return "";
}
finally i solved errors:
String input_path = "resources\\file.txt";
input_path = input_path.replace("\\", "/"); // doesn't work with back slash
URL file_url = getClass().getClassLoader().getResource(input_path);
String file_path = new URI(file_url.toString().replace(" ","%20")).getSchemeSpecificPart();
InputStream file_inputStream = file_url.openStream();
You can use class loader which will read from classpath as ROOT path (without "/" in the beginning)
InputStream in = getClass().getClassLoader().getResourceAsStream("file.txt");
BufferedReader reader = new BufferedReader(new InputStreamReader(in));
For some reason classLoader.getResource() always returned null when I deployed the web application to WildFly 14. getting classLoader from getClass().getClassLoader() or Thread.currentThread().getContextClassLoader() returns null.
getClass().getClassLoader() API doc says,
"Returns the class loader for the class. Some implementations may use null to represent the bootstrap class loader. This method will return null in such implementations if this class was loaded by the bootstrap class loader."
may be if you are using WildFly and yours web application try this
request.getServletContext().getResource() returned the resource url. Here request is an object of ServletRequest.
If you are using spring, then you can use the the following method to read file from src/main/resources:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import org.springframework.core.io.ClassPathResource;
public String readFileToString(String path) throws IOException {
StringBuilder resultBuilder = new StringBuilder("");
ClassPathResource resource = new ClassPathResource(path);
try (
InputStream inputStream = resource.getInputStream();
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream))) {
String line;
while ((line = bufferedReader.readLine()) != null) {
resultBuilder.append(line);
}
}
return resultBuilder.toString();
}
Below code works with Spring boot(kotlin):
val authReader = InputStreamReader(javaClass.getResourceAsStream("/file1.json"))
If you wanna read as a file, I believe there still is a similar solution:
ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("file/test.xml").getFile());
I would like to read a resource from within my jar like so:
File file;
file = new File(getClass().getResource("/file.txt").toURI());
BufferedReader reader = new BufferedReader(new FileReader(file));
//Read the file
and it works fine when running it in Eclipse, but if I export it to a jar, and then run it, there is an IllegalArgumentException:
Exception in thread "Thread-2"
java.lang.IllegalArgumentException: URI is not hierarchical
and I really don't know why but with some testing I found if I change
file = new File(getClass().getResource("/file.txt").toURI());
to
file = new File(getClass().getResource("/folder/file.txt").toURI());
then it works the opposite (it works in jar but not eclipse).
I'm using Eclipse and the folder with my file is in a class folder.
Rather than trying to address the resource as a File just ask the ClassLoader to return an InputStream for the resource instead via getResourceAsStream:
try (InputStream in = getClass().getResourceAsStream("/file.txt");
BufferedReader reader = new BufferedReader(new InputStreamReader(in))) {
// Use resource
}
As long as the file.txt resource is available on the classpath then this approach will work the same way regardless of whether the file.txt resource is in a classes/ directory or inside a jar.
The URI is not hierarchical occurs because the URI for a resource within a jar file is going to look something like this: file:/example.jar!/file.txt. You cannot read the entries within a jar (a zip file) like it was a plain old File.
This is explained well by the answers to:
How do I read a resource file from a Java jar file?
Java Jar file: use resource errors: URI is not hierarchical
To access a file in a jar you have two options:
Place the file in directory structure matching your package name (after extracting .jar file, it should be in the same directory as .class file), then access it using getClass().getResourceAsStream("file.txt")
Place the file at the root (after extracting .jar file, it should be in the root), then access it using Thread.currentThread().getContextClassLoader().getResourceAsStream("file.txt")
The first option may not work when jar is used as a plugin.
I had this problem before and I made fallback way for loading. Basically first way work within .jar file and second way works within eclipse or other IDE.
public class MyClass {
public static InputStream accessFile() {
String resource = "my-file-located-in-resources.txt";
// this is the path within the jar file
InputStream input = MyClass.class.getResourceAsStream("/resources/" + resource);
if (input == null) {
// this is how we load file within editor (eg eclipse)
input = MyClass.class.getClassLoader().getResourceAsStream(resource);
}
return input;
}
}
Up until now (December 2017), this is the only solution I found which works both inside and outside the IDE.
Use PathMatchingResourcePatternResolver
Note: it works also in spring-boot
In this example I'm reading some files located in src/main/resources/my_folder:
try {
// Get all the files under this inner resource folder: my_folder
String scannedPackage = "my_folder/*";
PathMatchingResourcePatternResolver scanner = new PathMatchingResourcePatternResolver();
Resource[] resources = scanner.getResources(scannedPackage);
if (resources == null || resources.length == 0)
log.warn("Warning: could not find any resources in this scanned package: " + scannedPackage);
else {
for (Resource resource : resources) {
log.info(resource.getFilename());
// Read the file content (I used BufferedReader, but there are other solutions for that):
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(resource.getInputStream()));
String line = null;
while ((line = bufferedReader.readLine()) != null) {
// ...
// ...
}
bufferedReader.close();
}
}
} catch (Exception e) {
throw new Exception("Failed to read the resources folder: " + e.getMessage(), e);
}
The problem is that certain third party libraries require file pathnames rather than input streams. Most of the answers don't address this issue.
In this case, one workaround is to copy the resource contents into a temporary file. The following example uses jUnit's TemporaryFolder.
private List<String> decomposePath(String path){
List<String> reversed = Lists.newArrayList();
File currFile = new File(path);
while(currFile != null){
reversed.add(currFile.getName());
currFile = currFile.getParentFile();
}
return Lists.reverse(reversed);
}
private String writeResourceToFile(String resourceName) throws IOException {
ClassLoader loader = getClass().getClassLoader();
InputStream configStream = loader.getResourceAsStream(resourceName);
List<String> pathComponents = decomposePath(resourceName);
folder.newFolder(pathComponents.subList(0, pathComponents.size() - 1).toArray(new String[0]));
File tmpFile = folder.newFile(resourceName);
Files.copy(configStream, tmpFile.toPath(), REPLACE_EXISTING);
return tmpFile.getAbsolutePath();
}
In my case I finally made it with
import java.lang.Thread;
import java.io.BufferedReader;
import java.io.InputStreamReader;
final BufferedReader in = new BufferedReader(new InputStreamReader(
Thread.currentThread().getContextClassLoader().getResourceAsStream("file.txt"))
); // no initial slash in file.txt
Make sure that you work with the correct separator. I replaced all / in a relative path with a File.separator. This worked fine in the IDE, however did not work in the build JAR.
I have found a fix
BufferedReader br = new BufferedReader(new InputStreamReader(Main.class.getResourceAsStream(path)));
Replace "Main" with the java class you coded it in. replace "path" with the path within the jar file.
for example, if you put State1.txt in the package com.issac.state, then type the path as "/com/issac/state/State1" if you run Linux or Mac. If you run Windows then type the path as "\com\issac\state\State1". Don't add the .txt extension to the file unless the File not found exception occurs.
This code works both in Eclipse and in Exported Runnable JAR
private String writeResourceToFile(String resourceName) throws IOException {
File outFile = new File(certPath + File.separator + resourceName);
if (outFile.isFile())
return outFile.getAbsolutePath();
InputStream resourceStream = null;
// Java: In caso di JAR dentro il JAR applicativo
URLClassLoader urlClassLoader = (URLClassLoader)Cypher.class.getClassLoader();
URL url = urlClassLoader.findResource(resourceName);
if (url != null) {
URLConnection conn = url.openConnection();
if (conn != null) {
resourceStream = conn.getInputStream();
}
}
if (resourceStream != null) {
Files.copy(resourceStream, outFile.toPath(), StandardCopyOption.REPLACE_EXISTING);
return outFile.getAbsolutePath();
} else {
System.out.println("Embedded Resource " + resourceName + " not found.");
}
return "";
}
finally i solved errors:
String input_path = "resources\\file.txt";
input_path = input_path.replace("\\", "/"); // doesn't work with back slash
URL file_url = getClass().getClassLoader().getResource(input_path);
String file_path = new URI(file_url.toString().replace(" ","%20")).getSchemeSpecificPart();
InputStream file_inputStream = file_url.openStream();
You can use class loader which will read from classpath as ROOT path (without "/" in the beginning)
InputStream in = getClass().getClassLoader().getResourceAsStream("file.txt");
BufferedReader reader = new BufferedReader(new InputStreamReader(in));
For some reason classLoader.getResource() always returned null when I deployed the web application to WildFly 14. getting classLoader from getClass().getClassLoader() or Thread.currentThread().getContextClassLoader() returns null.
getClass().getClassLoader() API doc says,
"Returns the class loader for the class. Some implementations may use null to represent the bootstrap class loader. This method will return null in such implementations if this class was loaded by the bootstrap class loader."
may be if you are using WildFly and yours web application try this
request.getServletContext().getResource() returned the resource url. Here request is an object of ServletRequest.
If you are using spring, then you can use the the following method to read file from src/main/resources:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import org.springframework.core.io.ClassPathResource;
public String readFileToString(String path) throws IOException {
StringBuilder resultBuilder = new StringBuilder("");
ClassPathResource resource = new ClassPathResource(path);
try (
InputStream inputStream = resource.getInputStream();
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream))) {
String line;
while ((line = bufferedReader.readLine()) != null) {
resultBuilder.append(line);
}
}
return resultBuilder.toString();
}
Below code works with Spring boot(kotlin):
val authReader = InputStreamReader(javaClass.getResourceAsStream("/file1.json"))
If you wanna read as a file, I believe there still is a similar solution:
ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("file/test.xml").getFile());
I have the following code trying to read a properties file:
Properties prop = new Properties();
ClassLoader loader = Thread.currentThread().getContextClassLoader();
InputStream stream = loader.getResourceAsStream("myProp.properties");
prop.load(stream);
I get an exception at the last line. Specifically:
Exception in thread "main" java.lang.NullPointerException
at java.util.Properties$LineReader.readLine(Properties.java:418)
at java.util.Properties.load0(Properties.java:337)
at java.util.Properties.load(Properties.java:325)
at Assignment1.BaseStation.readPropertyFile(BaseStation.java:46)
at Assignment1.BaseStation.main(BaseStation.java:87)
thanks,
Nikos
Based on your exception, the InputStream is null, this means the class loader is not finding your properties file. I'm guessing that myProp.properties is in the root of your project, if that's the case, you need a preceding slash:
InputStream stream = loader.getResourceAsStream("/myProp.properties");
You can find information on this page:
http://www.mkyong.com/java/java-properties-file-examples/
Properties prop = new Properties();
try {
//load a properties file from class path, inside static method
prop.load(App.class.getClassLoader().getResourceAsStream("config.properties"));
//get the property value and print it out
System.out.println(prop.getProperty("database"));
System.out.println(prop.getProperty("dbuser"));
System.out.println(prop.getProperty("dbpassword"));
}
catch (IOException ex) {
ex.printStackTrace();
}
You can use ResourceBundle class to read the properties file.
ResourceBundle rb = ResourceBundle.getBundle("myProp.properties");
Properties prop = new Properties();
try {
prop.load(new FileInputStream("conf/filename.properties"));
} catch (IOException e) {
e.printStackTrace();
}
conf/filename.properties base on project root dir
You can't use this keyword like -
props.load(this.getClass().getResourceAsStream("myProps.properties"));
in a static context.
The best thing would be to get hold of application context like -
ApplicationContext context = new ClassPathXmlApplicationContext("classpath:/META-INF/spring/app-context.xml");
then you can load the resource file from the classpath -
//load a properties file from class path, inside static method
prop.load(context.getClassLoader().getResourceAsStream("config.properties"));
This will work for both static and non static context and the best part is this properties file can be in any package/folder included in the application's classpath.
Your file should be available as com/example/foo/myProps.properties in classpath. Then load it as:
props.load(this.getClass().getResourceAsStream("myProps.properties"));
if your config.properties is not in src/main/resource directory and it is in root directory of the project then you need to do somethinglike below :-
Properties prop = new Properties();
File configFile = new File(myProp.properties);
InputStream stream = new FileInputStream(configFile);
prop.load(stream);
I see that the question is an old one. If anyone stumbles upon this in the future, I think this is one simple way of doing it.
Keep the properties file in your project folder.
FileReader reader = new FileReader("Config.properties");
Properties prop = new Properties();
prop.load(reader);
If your properties file path and your java class path are same then you should this.
For example:
src/myPackage/MyClass.java
src/myPackage/MyFile.properties
Properties prop = new Properties();
InputStream stream = MyClass.class.getResourceAsStream("MyFile.properties");
prop.load(stream);
Make sure that the file name is correct and that the file is actually in the class path. getResourceAsStream() will return null if this is not the case which causes the last line to throw the exception.
If myProp.properties is in the root directory of your project, use /myProp.properties instead.
You can use java.io.InputStream to read the file as shown below:
InputStream inputStream = getClass().getClassLoader().getResourceAsStream(myProps.properties);
Given the context loader.getResourceAsStream("myPackage/myProp.properties") should be used.
Leading '/' doesn't work with ClassLoader.getResourceAsStream(String) method.
Alternatively you could use Class.getResourceAsStream(String) method, which uses '/' to determine if the path is absolute or relative to the class location.
Examples:
myClass.class.getResourceAsStream("myProp.properties")
myClass.class.getResourceAsStream("/myPackage/myProp.properties")
Many answers here describe dangerous methods where they instantiate a file input stream but do not get a reference to the input stream in order to close the stream later. This results in dangling input streams and memory leaks. The correct way of loading the properties should be similar to following:
Properties prop = new Properties();
try(InputStream fis = new FileInputStream("myProp.properties")) {
prop.load(fis);
}
catch(Exception e) {
System.out.println("Unable to find the specified properties file");
e.printStackTrace();
return;
}
Note the instantiating of the file input stream in try-with-resources block. Since a FileInputStream is autocloseable, it will be automatically closed after the try-with-resources block is exited. If you want to use a simple try block, you must explicitly close it using fis.close(); in the finally block.
None of the current answers show the InputStream being closed (this will leak a file descriptor), and/or don't deal with .getResourceAsStream() returning null when the resource is not found (this will lead to a NullPointerException with the confusing message, "inStream parameter is null"). You need something like the following:
String propertiesFilename = "server.properties";
Properties prop = new Properties();
try (var inputStream = getClass().getClassLoader().getResourceAsStream(propertiesFilename)) {
if (inputStream == null) {
throw new FileNotFoundException(propertiesFilename);
}
prop.load(inputStream);
} catch (IOException e) {
throw new RuntimeException(
"Could not read " + propertiesFilename + " resource file: " + e);
}
A good practice which is not state in previous solution is to passing properties especially the property files that generated in compile time with build plugins perhaps, is to Use PropertySourcesPlaceholderConfigurer
#Bean
public static PropertySourcesPlaceholderConfigurer placeholderConfigurer() {
PropertySourcesPlaceholderConfigurer propsConfig
= new PropertySourcesPlaceholderConfigurer();
propsConfig.setLocation(new ClassPathResource("myProp.properties"));
propsConfig.setIgnoreResourceNotFound(true);
propsConfig.setIgnoreUnresolvablePlaceholders(true);
return propsConfig;
}
then can accessing the properties from IOC as demand such
#Value("${your.desired.property.pointer}")
private String value;
For Reading Properties file with its original order:
File file = new File("../config/edc.properties");
PropertiesConfiguration config = new PropertiesConfiguration();
PropertiesConfigurationLayout layout = new PropertiesConfigurationLayout(config);
layout.load(new InputStreamReader(new FileInputStream(file)));
for(Object propKey : layout.getKeys()){
PropertiesConfiguration propval = layout.getConfiguration();
String value = propval.getProperty((String) propKey).toString();
out.print("Current Key:" + propkey + "Current Value:" + propval + "<br>");
}
Specify the path starting from src as below:
src/main/resources/myprop.proper