I'm having trouble reading a text file from inside a jar. I managed it while it was unzipped but now that I've archived it it won't work.
The jar has:
BTA.jar
Splash.class
*.class
Splash.txt
etc..
Now, at the moment I'm using a BufferedReader and a FileReader
f = new FileReader("Splash.txt");
in = new BufferedReader(f);
but from googling around I know that that only reads files from outside of the jar.
My question is: How do I read a file from inside the jar?
Assuming you're executing the BTA.jar, you can use
InputStream in = YourClass.class.getResourceAsStream("/Splash.txt");
to retrieve an InputStream and pass it to the BufferedReader through an InputStreamReader.
The rules for the format of the path to pass to the method are defined here, in the javadoc for the Class class.
The problem is Splash.txt isn't a file, its series of bytes within in a (lightly) compressed file which has a named entry within a table of contents as specified by the Jar/Zip standard.
This measn you can use FileReader to read the file, instead, you need to access it via the Class resource management API, for example..
getClass().getResourceAsStream("/Splash.txt");
Or, if you're trying to read it from static context
Splash.class.getResourceAsStream("/Splash.txt");
Don't forget, you become responsible for closing the InputStream when you are done, for example...
try (InputStream is = getClass().getResourceAsStream("/Splash.txt")) {....
A jar is just a zip file and can be read like this..
ZipFile file = new ZipFile("/home/Desktop/myjar.jar");
Enumeration<? extends ZipEntry> entries = file.entries();
while (entries.hasMoreElements()) {
ZipEntry entry = entries.nextElement();
if (entry.getName().startsWith("splash")) {
try (BufferedReader reader = new BufferedReader(new InputStreamReader(file.getInputStream(entry)))) {
String line = null;
while ((line = reader.readLine()) != null) {
System.out.println(line);
}
}
}
}
Related
I have a tar file and which contains many files. I need to get a specific file from tar file and read data from that file.
I am untaring file using the below code and I will read this returned input stream using some other function.
private InputStream unTar(final File inputFile, final File outputDir) throws FileNotFoundException, IOException, ArchiveException {
InputStream versionInputStream = null;
final InputStream is = new FileInputStream(inputFile);
final TarArchiveInputStream debInputStream = (TarArchiveInputStream) new ArchiveStreamFactory().createArchiveInputStream("tar", is);
TarArchiveEntry entry = null;
while ((entry = (TarArchiveEntry)debInputStream.getNextEntry()) != null) {
if (!entry.isDirectory() && entry.getName().equals("version.txt")) {
versionInputStream = new FileInputStream(entry.getFile());
}
}
return versionInputStream;
}
I get null pointer exception when i do versionInputStream = new FileInputStream(entry.getFile());
I know that we can first save this file in directory and then read the file but i dont want to save this file in directory.
Is there some way I can read this file without saving the file to some dir?
There is no file for an entry of an archive you read. TarArchiveEntry's getFile method only returns anything useful when the entry has been created with a File-arg constructor, which only makes sense when creating an archive not reading it.
The stream you are looking for is the TarArchiveInputStream itself after you've positioned it at the entry you want to read, i.e.
if (!entry.isDirectory() && entry.getName().equals("version.txt")) {
versionInputStream = debInputStream;
break;
}
note the break.
The not-yet-released (an no, no release date, yet) Commons Compress 1.21 will contain a new TarFile class that provides random-access to archives read from a seekable source (like a File) and will make your task more convenient.
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 a properties file contains the file name only say file=fileName.dat. I've put the properties file under the class path and could read the file name(file.dat) properly from it in the mainClass. After reading the file name I passed the file name(just name not the path) to another class under a package say pack.myClass to read that file. But the problem is pack.myClass could not get the file path properly. I've put the file fileName.dat both inside and outside the packagepack but couldn't make it work.
Can anybody suggest me that where to put the file fileName.dat so I can read it properly and the whole application would be portable too.
Thanks!
The code I'm using to read the config file and getting the file name:
Properties prop = new Properties();
InputStream in = mainClass.class.getResourceAsStream("config.properties");
prop.load(in);
in.close();
myClass mc = new myClass();
mc.readTheFile(prop.getProperty("file"));
/*until this code is working good*/
Then in myClass which is under package named pack I am doing:
public void readTheFile(String filename) throws IOException {
FileReader fileReader = new FileReader(filename); /*this couldn't get the file whether i'm putting the file inside or outside the package folder */
/*after reading the file I've to do the BufferReader for further operation*/
BufferedReader bufferedReader = new BufferedReader(fileReader);
I assume that you are trying to read properties file using getResource method of class. If you put properties file on root of the classpath you should prefix file name with '/' to indicate root of classpath, for example getResource("/file.dat"). If properties file is under the same folder with the class you on which you invoke getResource method, than you should not use '/' prefix.
When you use a relative file name such as fileName.dat, you're asking for a file with this name in the current directory. The current directory has nothing to do with packages. It's the directory from which the JVM is started.
So if you're in the directory c:\foo\bar when you launch your application (using java -cp ... pack.MyClass), it will look for the file c:\foo\bar\fileName.dat.
Try..
myClass mc = new myClass();
InputStream in = mc.getClass().getResourceAsStream("/pack/config.properties");
..or simply
InputStream in = mc.getClass().getResourceAsStream("config.properties");
..for the last line if the main is in myClass The class loader available in the main() will often be the bootstrap class-loader, as opposed to the class-loader intended for application resources.
Class.getResource will look in your package directory for a file of the specified name.
JavaDocs here
Or getResourceAsStream is sometimes more convenient as you probably want to read the contents of the resource.
Most of the time it would be best to look for the "fileName.dat" somewhere in the "user.home" folder, which is a system property. First create a File path from the "user.home" and then try to find the file there. This is a bit of a guess as you don't provide the exact user of the application, but this would be the most common place.
You are currently reading from the current folder which is determined by
String currentDir = new File(".").getAbsolutePath();
or
System.getProperty("user.dir")
To read a file, even from within a jar archive:
readTheFile(String package, String filename) throws MalformedURLException, IOException
{
String filepath = package+"/"+filename;
// like "pack/fileName.dat" or "fileName.dat"
String s = (new SourceBase()).getSourceBase() + filepath;
URL url = new URL(s);
InputStream ins = url.openStream();
BufferedReader rdr = new BufferedReader(new InputStreamReader(ins, "utf8"));
do {
s = rdr.readLine();
if(s!= null) System.out.println(s);
}
while(s!=null);
rdr.close();
}
with
class SourceBase
{
public String getSourceBase()
{
String cn = this.getClass().getName().replace('.', '/') + ".class";
// like "packagex/SourceBase.class"
String s = this.getClass().getResource('/' + cn).toExternalForm();
// like "file:/javadir/Projects/projectX/build/classes/packagex/SourceBase.class"
// or "jar:file:/opt/java/PROJECTS/testProject/dist/
// testProject.jar!/px/SourceBase.class"
return s.substring(0, s.lastIndexOf(cn));
// like "file:/javadir/Projects/projectX/build/classes/"
// or "jar:file:/opt/java/PROJECTS/testProject/dist/testProject.jar!/"
}
}