I'm trying to access a resource from a jar file. The resource is located in the same directory where is the jar.
my-dir:
tester.jar
test.jpg
I tried different things including the following, but every time the input stream is null:
[1]
String path = new File(".").getAbsolutePath();
InputStream inputStream = this.getClass().getResourceAsStream(path.replace("\\.", "\\") + "test.jpg");
[2]
File f = new File(this.getClass().getProtectionDomain().getCodeSource().getLocation().toURI().getPath());
InputStream inputStream = this.getClass().getResourceAsStream(f.getParent() + "test.jpg");
Can you give me some hints? Thanks.
If you are sure, that your application's current folder is the folder of the jar, you can simply call InputStream f = new FileInputStream("test.jpg");
The getResource methods will load stuff using the classloader, not through filesystem. This is why your approach (1) failed.
If the folder containing your *.jar and image file is in the classpath, you can get the image resource as if it was on the default-package:
class.getClass().getResourceAsStream("/test.jpg");
Beware: The image is now loaded in the classloader, and as long as the application runs, the image is not unloaded and served from memory if you load it again.
If the path containing the jar file is not given in the classpath, your approach to get the jarfile path is good.
But then simply access the file directly through the URI, by opening a stream on it:
URL u = this.getClass().getProtectionDomain().getCodeSource().getLocation();
// u2 is the url derived from the codesource location
InputStream s = u2.openStream();
Use this tutorial to help you create a URL to a single file in a jar file.
Here's an example:
String jarPath = "/home/user/myJar.jar";
String urlStr = "jar:file://" + jarPath + "!/test.jpg";
InputStream is = null;
try {
URL url = new URL(urlStr);
is = url.openStream();
Image image = ImageIO.read(is);
}
catch(Exception e) {
e.printStackTrace();
}
finally {
try {
is.close();
} catch(Exception IGNORE) {}
}
Related
I would like to load an image from my current src directory where the java class files are located as well. However, I always get an IOException..
And how can I make sure the file gets loaded properly on Mac/Linux as well on Windows?
My code so far:
String dir = System.getProperty("user.dir") + "/Logo_transparent.png";
File imageFile = new File(dir);
BufferedImage bufferedImage = null;
try {
bufferedImage = ImageIO.read(imageFile);
} catch (IOException e) {
System.out.println(e.getMessage());
System.out.println(dir);
System.out.println();
}
IOException message:
Can't read input file!
(My path is correct - is it because of the space between Google and Drive?)
/Users/myMac/Google Drive/Privat/Programming/Logo_transparent.png
Kind regards and thank you!
I think It's because you didn't create the file, You can create the file if it doesn't exist by using this code
if(!imageFile.exists()) imageFile.createNewFile();
You're code will look like this
String dir = System.getProperty("user.dir") + "/Logo_transparent.png";
File imageFile = new File(dir);
BufferedImage bufferedImage = null;
try {
if(!imageFile.exists()) imageFile.createNewFile();
bufferedImage = ImageIO.read(imageFile);
} catch (IOException e) {
System.out.println(e.getMessage());
System.out.println(dir);
System.out.println();
}
Also you shouldn't concat child files like that instead pass it as a second argument.
File imageFile = new File(System.getProperty("user.dir"), "Logo_transparent.png");
If your image files will be packaged together with your class files (for example in the same .jar) you should not use File but read it as a resource:
bufferedImage = ImageIO.read(this.getClass().getResourceAsStream("/Logo_transparent.png"));
Notice the '/' before the file name. This means to search in the root path of the classpath.
If you specify without / it will search in the package of this (the current class)
this.getClass().getResourceAsStream("Logo_transparent.png")
You can try to build the absolute path to the image like here and read it afterward.
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'm trying to access a resource from a jar file. The resource is located in the same directory where is the jar.
my-dir:
tester.jar
test.jpg
I tried different things including the following, but every time the input stream is null:
[1]
String path = new File(".").getAbsolutePath();
InputStream inputStream = this.getClass().getResourceAsStream(path.replace("\\.", "\\") + "test.jpg");
[2]
File f = new File(this.getClass().getProtectionDomain().getCodeSource().getLocation().toURI().getPath());
InputStream inputStream = this.getClass().getResourceAsStream(f.getParent() + "test.jpg");
Can you give me some hints? Thanks.
If you are sure, that your application's current folder is the folder of the jar, you can simply call InputStream f = new FileInputStream("test.jpg");
The getResource methods will load stuff using the classloader, not through filesystem. This is why your approach (1) failed.
If the folder containing your *.jar and image file is in the classpath, you can get the image resource as if it was on the default-package:
class.getClass().getResourceAsStream("/test.jpg");
Beware: The image is now loaded in the classloader, and as long as the application runs, the image is not unloaded and served from memory if you load it again.
If the path containing the jar file is not given in the classpath, your approach to get the jarfile path is good.
But then simply access the file directly through the URI, by opening a stream on it:
URL u = this.getClass().getProtectionDomain().getCodeSource().getLocation();
// u2 is the url derived from the codesource location
InputStream s = u2.openStream();
Use this tutorial to help you create a URL to a single file in a jar file.
Here's an example:
String jarPath = "/home/user/myJar.jar";
String urlStr = "jar:file://" + jarPath + "!/test.jpg";
InputStream is = null;
try {
URL url = new URL(urlStr);
is = url.openStream();
Image image = ImageIO.read(is);
}
catch(Exception e) {
e.printStackTrace();
}
finally {
try {
is.close();
} catch(Exception IGNORE) {}
}
I want to programmatically access a specific file which will be included in my project folder. Is there a way to do this? If so, where in my project folder do I put the file, and what is some simple code to get its file path?
private void saveFileToDrive() {
Thread t = new Thread(new Runnable() {
#Override
public void run() {
try {
java.io.File spreadsheet = new java.io.File("Untitled spreadsheet.xlsx");
String filePath = spreadsheet.getAbsolutePath();
System.out.println("file path is"+filePath);
URL fileURL = getClass().getClassLoader().getResource("Untitled spreadsheet.xlsx");
String filePath2 = fileURL.getPath();
System.out.println("file path2 is"+filePath2);
java.io.File fileContent = new java.io.File(filePath);
FileContent mediaContent = new FileContent("application/vnd.ms-excel", fileContent);
File body = new File();
body.setTitle(fileContent.getName());
body.setMimeType("application/vnd.ms-excel");
File file = service.files().insert(body, mediaContent).setConvert(true).execute();
if (file != null) {
showToast("File uploaded: " + file.getTitle());
}
else
;
} catch (UserRecoverableAuthIOException e) {
startActivityForResult(e.getIntent(), REQUEST_AUTHORIZATION);
} catch (IOException e) {
e.printStackTrace();
}
}
});
t.start();
}
Put the file in root folder of your project. Then get the File URL, Path and other details as:
File file = new File("test.txt");
String filePath = file.getAbsolutePath();
EDIT: Alternate way (if the file is in your classpath e.g. put the file in "src" folder, and make sure its moved in "bin" or "classes" folder after compilation):
URL fileURL = getClass().getClassLoader().getResource(fileName);
String fileName = fileURL.getFile();
String filePath = fileURL.getPath();
This depends a lot on what type of file you want to access. You can put the file in either assets or an appropriate subdirectory of res (see Difference between /res and /assets directories).
So you want to access a file internal to your app; and you want to do so directly, rather, that is, from an Android Context (and then with a [android.|<package_name>.]R.<resource_type>.<resource_name>).
You have two choices as to location: the res/raw folder or assets/ folder (outside of the res parent).
To choose between the two note from https://developer.android.com/guide/topics/resources/providing-resources.html
Arbitrary files to save in their raw form. To open these resources with a raw InputStream, call Resources.openRawResource() with the resource ID, which is R.raw.filename.
However, if you need access to original file names and file hierarchy, you might consider saving some resources in the assets/ directory (instead of res/raw/). Files in assets/ aren't given a resource ID, so you can read them only using AssetManager.
To access a file in res/raw/ directly rather, that is, from an Android Context (and then with a [android.|<package_name>.]R.<resource_type>.<resource_name>) you can do something like this:
File file = new File("app/src/main/res/raw/country_data_from_world_bank.xml");
BufferedReader bufferedReader = new BufferedReader(new FileReader(file));
If I have a resource on a classpath, I can both load it as stream fine, and there is even a URL representation of it. Unfortunately some implementations of the Url do not implement lastModified correctly.
What I would like is to take a path to something in the classpath, and then resolve it to a file that it is in on disk - if it in a jar, then a File pointing to the jar is fine. I can then get the lastModified from the File object instead of the URL, which will be more helpful.
Roughly speaking:
URL url = this.getClass().getResource(myResource);
String fileName;
if (url.getProtocol().equals("file")) {
fileName = url.getFile();
} else if (url.getProtocol().equals("jar")) {
JarURLConnection jarUrl = (JarURLConnection) url.openConnection();
fileName = jarUrl.getJarFile().getName();
} else {
throw new IllegalArgumentException("Not a file");
}
File file = new File(fileName);
long lastModified = file.lastModified();
Should do what you want. You will need to catch IOException.
No. This can't be done generally because URL can represent resources which are not associated with a file. For example, it can be HTTP, FTP or JNDI etc.
You can check for protocol and create the File yourself if the protocol is file-based, like "file://path", "jar://path!...".