Reading a resource file from within jar

问题:

I would like to read a resource from within my jar like so:

File file;
file = new File(getClass().getResource("/file.txt").toURI());
BufferredReader 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 the 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.

回答1:

Rather than trying to address the resource as a File just ask the ClassLoader to return an InputStream for the resource instead via getResourceAsStream:

InputStream in = getClass().getResourceAsStream("/file.txt"); 
BufferedReader reader = new BufferedReader(new InputStreamReader(in));

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

回答2:

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.

回答3:

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());

回答4:

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;
    }
}

回答5:

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.

回答6:

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);
}

回答7:

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();
  }

回答8:

After a lot of digging around in Java, the only solution that seems to work for me is to manually read the jar file itself unless you're in a development environment(IDE):

/** @return The root folder or jar file that the class loader loaded from */
public static final File getClasspathFile() {
    return new File(YourMainClass.class.getProtectionDomain().getCodeSource().getLocation().getFile());
}

/** @param resource The path to the resource
 * @return An InputStream containing the resource's contents, or
 *         <b><code>null</code></b> if the resource does not exist */
public static final InputStream getResourceAsStream(String resource) {
    resource = resource.startsWith("/") ? resource : "/" + resource;
    if(getClasspathFile().isDirectory()) {//Development environment:
        return YourMainClass.class.getResourceAsStream(resource);
    }
    final String res = resource;//Jar or exe:
    return AccessController.doPrivileged(new PrivilegedAction<InputStream>() {
        @SuppressWarnings("resource")
        @Override
        public InputStream run() {
            try {
                final JarFile jar = new JarFile(getClasspathFile());
                String resource = res.startsWith("/") ? res.substring(1) : res;
                if(resource.endsWith("/")) {//Directory; list direct contents:(Mimics normal getResourceAsStream("someFolder/") behaviour)
                    ByteArrayOutputStream baos = new ByteArrayOutputStream();
                    Enumeration<JarEntry> entries = jar.entries();
                    while(entries.hasMoreElements()) {
                        JarEntry entry = entries.nextElement();
                        if(entry.getName().startsWith(resource) && entry.getName().length() > resource.length()) {
                            String name = entry.getName().substring(resource.length());
                            if(name.contains("/") ? (name.endsWith("/") && (name.indexOf("/") == name.lastIndexOf("/"))) : true) {//If it's a folder, we don't want the children's folders, only the parent folder's children!
                                name = name.endsWith("/") ? name.substring(0, name.length() - 1) : name;
                                baos.write(name.getBytes(StandardCharsets.UTF_8));
                                baos.write('\r');
                                baos.write('\n');
                            }
                        }
                    }
                    jar.close();
                    return new ByteArrayInputStream(baos.toByteArray());
                }
                JarEntry entry = jar.getJarEntry(resource);
                InputStream in = entry != null ? jar.getInputStream(entry) : null;
                if(in == null) {
                    jar.close();
                    return in;
                }
                final InputStream stream = in;//Don't manage 'jar' with try-with-resources or close jar until the
                return new InputStream() {//returned stream is closed(closing the jar closes all associated InputStreams):
                    @Override
                    public int read() throws IOException {
                        return stream.read();
                    }

                    @Override
                    public int read(byte b[]) throws IOException {
                        return stream.read(b);
                    }

                    @Override
                    public int read(byte b[], int off, int len) throws IOException {
                        return stream.read(b, off, len);
                    }

                    @Override
                    public long skip(long n) throws IOException {
                        return stream.skip(n);
                    }

                    @Override
                    public int available() throws IOException {
                        return stream.available();
                    }

                    @Override
                    public void close() throws IOException {
                        try {
                            jar.close();
                        } catch(IOException ignored) {
                        }
                        stream.close();
                    }

                    @Override
                    public synchronized void mark(int readlimit) {
                        stream.mark(readlimit);
                    }

                    @Override
                    public synchronized void reset() throws IOException {
                        stream.reset();
                    }

                    @Override
                    public boolean markSupported() {
                        return stream.markSupported();
                    }
                };
            } catch(Throwable e) {
                e.printStackTrace();
                return null;
            }
        }
    });
}

Note: The above code only seems to work correctly for jar files if it is in the main class. I'm not sure why.

回答9:

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));

回答10:

I think this should be works in java as well. The following code I use is using kotlin.

val resource = Thread.currentThread().contextClassLoader.getResource('resources.txt')

回答11:

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.

回答12:

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();
    }

回答13:

Below code works with Spring boot(kotlin):

val authReader = InputStreamReader(javaClass.getResourceAsStream("/file1.json"))