blocks|key|546538|text|此link将告诉您如何操作。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|546539|神奇的是getResourceAsStream()方法：|546540|InputStream+is+=+
this.getClass().getClassLoader().getResourceAsStream("yourpackage/mypackage/myfile.xml")|code-block|syntax|javascript|546541|entityMap|0|LINK|mutability|MUTABLE|url|http://www.jugpadova.it/articles/2006/02/05/accessing-a-resource-within-a-jar^0|1|4|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|6|7|W|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|X|8|@]|9|@]|C|$I|J]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

This <a href="http://www.jugpadova.it/articles/2006/02/05/accessing-a-resource-within-a-jar" rel="nofollow">link</a> tells you how.

The magic is the getResourceAsStream() method :

<pre><code>InputStream is = 
this.getClass().getClassLoader().getResourceAsStream("yourpackage/mypackage/myfile.xml")
</code></pre>

blocks|key|546601|text|尝试以下操作。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|546602|将资源路径设置为包，例如，如果您的类路径是com.abc.package.MyClass，并且您的资源文件位于src/com/abc/+"<PathRelativeToThisClassFile>/<ResourceDirectory>"+/resources/：|offset|length|style|CODE|546603|URL+url+=+MyClass.class.getResource("resources/");
if+(url+==+null)+{
+++++//+error+-+missing+folder
}+else+{
++++File+dir+=+new+File(url.toURI());
++++for+(File+nextFile+:+dir.listFiles())+{
++++++++//+Do+something+with+nextFile
++++}
}|code-block|syntax|javascript|546604|您还可以使用|546605|URL+url+=+MyClass.class.getResource("/com/abc/package/resources/");|546606|entityMap^0|0|1W|1F|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@$D|U|E|V|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|W|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|J|7|Y|8|@]|9|@]|A|$K|L]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Try the following. 
Make the resource path <code>"&lt;PathRelativeToThisClassFile&gt;/&lt;ResourceDirectory&gt;"</code> E.g. if your class path is com.abc.package.MyClass and your resoure files are within src/com/abc/package/resources/:

<pre><code>URL url = MyClass.class.getResource("resources/");
if (url == null) {
 // error - missing folder
} else {
 File dir = new File(url.toURI());
 for (File nextFile : dir.listFiles()) {
 // Do something with nextFile
 }
}
</code></pre>

You can also use 

<pre><code>URL url = MyClass.class.getResource("/com/abc/package/resources/");
</code></pre>

blocks|key|546655|text|最后，我找到了解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|546656|final+String+path+=+"sample/folder";
final+File+jarFile+=+new+File(getClass().getProtectionDomain().getCodeSource().getLocation().getPath());

if(jarFile.isFile())+{++//+Run+with+JAR+file
++++final+JarFile+jar+=+new+JarFile(jarFile);
++++final+Enumeration<JarEntry>+entries+=+jar.entries();+//gives+ALL+entries+in+jar
++++while(entries.hasMoreElements())+{
++++++++final+String+name+=+entries.nextElement().getName();
++++++++if+(name.startsWith(path+%2B+"/"))+{+//filter+according+to+the+path
++++++++++++System.out.println(name);
++++++++}
++++}
++++jar.close();
}+else+{+//+Run+with+IDE
++++final+URL+url+=+Launcher.class.getResource("/"+%2B+path);
++++if+(url+!=+null)+{
++++++++try+{
++++++++++++final+File+apps+=+new+File(url.toURI());
++++++++++++for+(File+app+:+apps.listFiles())+{
++++++++++++++++System.out.println(app);
++++++++++++}
++++++++}+catch+(URISyntaxException+ex)+{
++++++++++++//+never+happens
++++++++}
++++}
}|code-block|syntax|javascript|546657|当你在IDE+(而不是jar文件)上运行应用程序时，第二个代码块才起作用，如果你不喜欢它，你可以删除它。|546658|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Finally, I found the solution:

<pre><code>final String path = "sample/folder";
final File jarFile = new File(getClass().getProtectionDomain().getCodeSource().getLocation().getPath());

if(jarFile.isFile()) { // Run with JAR file
 final JarFile jar = new JarFile(jarFile);
 final Enumeration&lt;JarEntry&gt; entries = jar.entries(); //gives ALL entries in jar
 while(entries.hasMoreElements()) {
 final String name = entries.nextElement().getName();
 if (name.startsWith(path + "/")) { //filter according to the path
 System.out.println(name);
 }
 }
 jar.close();
} else { // Run with IDE
 final URL url = Launcher.class.getResource("/" + path);
 if (url != null) {
 try {
 final File apps = new File(url.toURI());
 for (File app : apps.listFiles()) {
 System.out.println(app);
 }
 } catch (URISyntaxException ex) {
 // never happens
 }
 }
}
</code></pre>

The second block just work when you run the application on IDE (not with jar file), You can remove it if you don't like that.

blocks|key|329865|text|很简单..。使用OSGi。在OSGi中，您可以使用findEntries和findPaths遍历包的条目。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|329866|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Simple ... use OSGi. In OSGi you can iterate over your Bundle's entries with findEntries and findPaths.

blocks|key|329925|text|在我的jar文件中，我有一个名为Upload的文件夹，这个文件夹中还有另外三个文本文件，我需要在jar文件之外有一个完全相同的文件夹和文件，我使用了以下代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|329926|URL+inputUrl+=+getClass().getResource("/upload/blabla1.txt");
File+dest1+=+new+File("upload/blabla1.txt");
FileUtils.copyURLToFile(inputUrl,+dest1);

URL+inputUrl2+=+getClass().getResource("/upload/blabla2.txt");
File+dest2+=+new+File("upload/blabla2.txt");
FileUtils.copyURLToFile(inputUrl2,+dest2);

URL+inputUrl3+=+getClass().getResource("/upload/blabla3.txt");
File+dest3+=+new+File("upload/Bblabla3.txt");
FileUtils.copyURLToFile(inputUrl3,+dest3);|code-block|syntax|javascript|329927|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Inside my jar file I had a folder called Upload, this folder had three other text files inside it and I needed to have an exactly the same folder and files outside of the jar file, I used the code below:

<pre><code>URL inputUrl = getClass().getResource("/upload/blabla1.txt");
File dest1 = new File("upload/blabla1.txt");
FileUtils.copyURLToFile(inputUrl, dest1);

URL inputUrl2 = getClass().getResource("/upload/blabla2.txt");
File dest2 = new File("upload/blabla2.txt");
FileUtils.copyURLToFile(inputUrl2, dest2);

URL inputUrl3 = getClass().getResource("/upload/blabla3.txt");
File dest3 = new File("upload/Bblabla3.txt");
FileUtils.copyURLToFile(inputUrl3, dest3);
</code></pre>

blocks|key|546802|text|我知道这是很多年前的事了。但仅供遇到此话题的其他人使用。您可以做的是对目录路径使用getResourceAsStream()方法，输入流将拥有该目录中的所有文件名。之后，您可以将目录路径与每个文件名连接在一起，并在循环中为每个文件调用getResourceAsStream。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|546803|entityMap^0|15|L|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

I know this is many years ago . But just for other people come across this topic.
What you could do is to use <code>getResourceAsStream()</code> method with the directory path, and the input Stream will have all the files name from that dir. After that you can concat the dir path with each file name and call getResourceAsStream for each file in a loop.

blocks|key|546855|text|当我试图从jar中打包的资源中加载一些hadoop配置时，我也遇到了同样的问题……在IDE和jar+(发布版本)上。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|546856|我发现在遍历本地文件系统和jar上的目录内容时，java.nio.file.DirectoryStream工作得最好。|offset|length|style|CODE|546857|String+fooFolder+=+"/foo/folder";
....

ClassLoader+classLoader+=+foofClass.class.getClassLoader();
try+{
++++uri+=+classLoader.getResource(fooFolder).toURI();
}+catch+(URISyntaxException+e)+{
++++throw+new+FooException(e.getMessage());
}+catch+(NullPointerException+e){
++++throw+new+FooException(e.getMessage());
}

if(uri+==+null){
++++throw+new+FooException("something+is+wrong+directory+or+files+missing");
}

/**+i+want+to+know+if+i+am+inside+the+jar+or+working+on+the+IDE*/
if(uri.getScheme().contains("jar")){
++++/**+jar+case+*/
++++try{
++++++++URL+jar+=+FooClass.class.getProtectionDomain().getCodeSource().getLocation();
++++++++//jar.toString()+begins+with+file:
++++++++//i+want+to+trim+it+out...
++++++++Path+jarFile+=+Paths.get(jar.toString().substring("file:".length()));
++++++++FileSystem+fs+=+FileSystems.newFileSystem(jarFile,+null);
++++++++DirectoryStream<Path>+directoryStream+=+Files.newDirectoryStream(fs.getPath(fooFolder));
++++++++for(Path+p:+directoryStream){
++++++++++++InputStream+is+=+FooClass.class.getResourceAsStream(p.toString())+;
++++++++performFooOverInputStream(is);
++++++++/**+your+logic+here+**/
++++++++++++}
++++}catch(IOException+e)+{
++++++++throw+new+FooException(e.getMessage());+++++
++++}
}
else{
++++/**+IDE+case+*/
++++Path+path+=+Paths.get(uri);
++++try+{
++++++++DirectoryStream<Path>+directoryStream+=+Files.newDirectoryStream(path);
++++++++for(Path+p+:+directoryStream){
++++++++++++InputStream+is+=+new+FileInputStream(p.toFile());
++++++++++++performFooOverInputStream(is);
++++++++}
++++}+catch+(IOException+_e)+{
++++++++throw+new+FooException(_e.getMessage());
++++}
}|code-block|syntax|javascript|546858|entityMap^0|0|O|T|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@$D|Q|E|R|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|S|8|@]|9|@]|A|$K|L]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

I had the same problem at hands while i was attempting to load some hadoop configurations from resources packed in the jar... on both the IDE and on jar (release version).

I found <code>java.nio.file.DirectoryStream</code> to work the best to iterate over directory contents over both local filesystem and jar.

<pre><code>String fooFolder = "/foo/folder";
....

ClassLoader classLoader = foofClass.class.getClassLoader();
try {
 uri = classLoader.getResource(fooFolder).toURI();
} catch (URISyntaxException e) {
 throw new FooException(e.getMessage());
} catch (NullPointerException e){
 throw new FooException(e.getMessage());
}

if(uri == null){
 throw new FooException("something is wrong directory or files missing");
}

/** i want to know if i am inside the jar or working on the IDE*/
if(uri.getScheme().contains("jar")){
 /** jar case */
 try{
 URL jar = FooClass.class.getProtectionDomain().getCodeSource().getLocation();
 //jar.toString() begins with file:
 //i want to trim it out...
 Path jarFile = Paths.get(jar.toString().substring("file:".length()));
 FileSystem fs = FileSystems.newFileSystem(jarFile, null);
 DirectoryStream&lt;Path&gt; directoryStream = Files.newDirectoryStream(fs.getPath(fooFolder));
 for(Path p: directoryStream){
 InputStream is = FooClass.class.getResourceAsStream(p.toString()) ;
 performFooOverInputStream(is);
 /** your logic here **/
 }
 }catch(IOException e) {
 throw new FooException(e.getMessage()); 
 }
}
else{
 /** IDE case */
 Path path = Paths.get(uri);
 try {
 DirectoryStream&lt;Path&gt; directoryStream = Files.newDirectoryStream(path);
 for(Path p : directoryStream){
 InputStream is = new FileInputStream(p.toFile());
 performFooOverInputStream(is);
 }
 } catch (IOException _e) {
 throw new FooException(_e.getMessage());
 }
}
</code></pre>

blocks|key|546890|text|另一种解决方案是，您可以像这样使用ResourceLoader完成此操作：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|546891|import+org.springframework.core.io.Resource;
import+org.apache.commons.io.FileUtils;

@Autowire
private+ResourceLoader+resourceLoader;

...

Resource+resource+=+resourceLoader.getResource("classpath:/path/to/you/dir");
File+file+=+resource.getFile();
Iterator<File>+fi+=+FileUtils.iterateFiles(file,+null,+true);
while(fi.hasNext())+{
++++load(fi.next())
}|code-block|syntax|javascript|546892|entityMap^0|H|E|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Another solution, you can do it using <code>ResourceLoader</code> like this:

<pre><code>import org.springframework.core.io.Resource;
import org.apache.commons.io.FileUtils;

@Autowire
private ResourceLoader resourceLoader;

...

Resource resource = resourceLoader.getResource("classpath:/path/to/you/dir");
File file = resource.getFile();
Iterator&lt;File&gt; fi = FileUtils.iterateFiles(file, null, true);
while(fi.hasNext()) {
 load(fi.next())
}
</code></pre>

blocks|key|330213|text|下面的代码将想要的“文件夹”作为路径返回，而不管它是否在jar中。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|330214|++private+Path+getFolderPath()+throws+URISyntaxException,+IOException+{
++++URI+uri+=+getClass().getClassLoader().getResource("folder").toURI();
++++if+("jar".equals(uri.getScheme()))+{
++++++FileSystem+fileSystem+=+FileSystems.newFileSystem(uri,+Collections.emptyMap(),+null);
++++++return+fileSystem.getPath("path/to/folder/inside/jar");
++++}+else+{
++++++return+Paths.get(uri);
++++}
++}|code-block|syntax|javascript|330215|需要java+7%2B。|330216|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

The following code returns the wanted "folder" as Path regardless of if it is inside a jar or not. 

<pre><code> private Path getFolderPath() throws URISyntaxException, IOException {
 URI uri = getClass().getClassLoader().getResource("folder").toURI();
 if ("jar".equals(uri.getScheme())) {
 FileSystem fileSystem = FileSystems.newFileSystem(uri, Collections.emptyMap(), null);
 return fileSystem.getPath("path/to/folder/inside/jar");
 } else {
 return Paths.get(uri);
 }
 }
</code></pre>

Requires java 7+.

blocks|key|547032|text|下面的代码从自定义资源目录中获取.yaml文件。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|547033|ClassLoader+classLoader+=+this.getClass().getClassLoader();
++++++++++++URI+uri+=+classLoader.getResource(directoryPath).toURI();
++++
++++++++++++if("jar".equalsIgnoreCase(uri.getScheme())){
++++++++++++++++Pattern+pattern+=+Pattern.compile("%5E.%2B"+%2B"/classes/"+%2B+directoryPath+%2B+"/.%2B.yaml$");
++++++++++++++++log.debug("pattern+{}+",+pattern.pattern());
++++++++++++++++ApplicationHome+home+=+new+ApplicationHome(SomeApplication.class);
++++++++++++++++JarFile+file+=+new+JarFile(home.getSource());
++++++++++++++++Enumeration<JarEntry>++jarEntries+=+file.entries()+;
++++++++++++++++while(jarEntries.hasMoreElements()){
++++++++++++++++++++JarEntry+entry+=+jarEntries.nextElement();
++++++++++++++++++++Matcher+matcher+=+pattern.matcher(entry.getName());
++++++++++++++++++++if(matcher.find()){
++++++++++++++++++++++++InputStream+in+=
++++++++++++++++++++++++++++++++file.getInputStream(entry);
++++++++++++++++++++++++//work+on+the+stream
++++
++++
++++++++++++++++++++}
++++++++++++++++}
++++
++++++++++++}else{
++++++++++++++++//When+Spring+boot+application+executed+through+Non-Jar+strategy+like+through+IDE+or+as+a+War.
++++
++++++++++++++++String+path+=+uri.getPath();
++++++++++++++++File[]+files+=+new+File(path).listFiles();
+++++++++++++++
++++++++++++++++for(File+file:+files){
++++++++++++++++++++if(file+!=+null){
++++++++++++++++++++++++try+{
++++++++++++++++++++++++++++InputStream+is+=+new+FileInputStream(file);
+++++++++++++++++++++++++++//work+on+stream
++++++++++++++++++++++++++++
++++++++++++++++++++++++}+catch+(Exception+e)+{
++++++++++++++++++++++++++++log.error("Exception+while+parsing+file+yaml+file+{}+:+{}+"+,+file.getAbsolutePath(),+e.getMessage());
++++++++++++++++++++++++}
++++++++++++++++++++}else{
++++++++++++++++++++++++log.warn("File+Object+is+null+while+parsing+yaml+file");
++++++++++++++++++++}
++++++++++++++++}
++++++++++++}|code-block|syntax|javascript|547034|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Below code gets .yaml files from a custom resource directory.
<pre><code>ClassLoader classLoader = this.getClass().getClassLoader();
 URI uri = classLoader.getResource(directoryPath).toURI();
 
 if(&quot;jar&quot;.equalsIgnoreCase(uri.getScheme())){
 Pattern pattern = Pattern.compile(&quot;^.+&quot; +&quot;/classes/&quot; + directoryPath + &quot;/.+.yaml$&quot;);
 log.debug(&quot;pattern {} &quot;, pattern.pattern());
 ApplicationHome home = new ApplicationHome(SomeApplication.class);
 JarFile file = new JarFile(home.getSource());
 Enumeration&lt;JarEntry&gt; jarEntries = file.entries() ;
 while(jarEntries.hasMoreElements()){
 JarEntry entry = jarEntries.nextElement();
 Matcher matcher = pattern.matcher(entry.getName());
 if(matcher.find()){
 InputStream in =
 file.getInputStream(entry);
 //work on the stream
 
 
 }
 }
 
 }else{
 //When Spring boot application executed through Non-Jar strategy like through IDE or as a War.
 
 String path = uri.getPath();
 File[] files = new File(path).listFiles();
 
 for(File file: files){
 if(file != null){
 try {
 InputStream is = new FileInputStream(file);
 //work on stream
 
 } catch (Exception e) {
 log.error(&quot;Exception while parsing file yaml file {} : {} &quot; , file.getAbsolutePath(), e.getMessage());
 }
 }else{
 log.warn(&quot;File Object is null while parsing yaml file&quot;);
 }
 }
 }
 
</code></pre>

blocks|key|547094|text|如果你使用的是Spring，你可以使用org.springframework.core.io.support.PathMatchingResourcePatternResolver来处理Resource对象而不是文件。当在Jar文件内部和外部运行时，这是有效的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|547095|PathMatchingResourcePatternResolver+r+=+new+PathMatchingResourcePatternResolver();
Resource[]+resources+=+r.getResources("/myfolder/*");|code-block|syntax|javascript|547096|然后，您可以使用getInputStream和getFilename中的文件名来访问数据。|547097|请注意，如果您在从Jar运行时尝试使用getFile，它仍然会失败。|547098|entityMap^0|J|1Z|2L|8|0|0|8|E|N|B|0|J|7|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|11|8|@$9|12|A|13|B|C]]|D|@]|E|$]]|$1|O|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|P|$]]

If you are using Spring you can use <code>org.springframework.core.io.support.PathMatchingResourcePatternResolver</code> and deal with <code>Resource</code> objects rather than files. This works when running inside and outside of a Jar file.
<pre><code>PathMatchingResourcePatternResolver r = new PathMatchingResourcePatternResolver();
Resource[] resources = r.getResources(&quot;/myfolder/*&quot;);
</code></pre>
Then you can access the data using <code>getInputStream</code> and the filename from <code>getFilename</code>.
Note that it will still fail if you try to use the <code>getFile</code> while running from a Jar.

I have a resources folder/package in the root of my project, I &quot;don't&quot; want to load a certain File. If I wanted to load a certain File, I would use class.getResourceAsStream and I would be fine!! What I actually want to do is to load a &quot;Folder&quot; within the resources folder, loop on the Files inside that Folder and get a Stream to each file and read in the content... Assume that the File names are not determined before runtime... What should I do? Is there a way to get a list of the files inside a Folder in your jar File?
Notice that the Jar file with the resources is the same jar file from which the code is being run...

How can I access a folder inside of a resource folder from inside my jar File?

文件系统

我在项目的根目录下有一个资源文件夹/包，我“不”想加载某个文件。如果我想加载某个文件，我会使用class.getResourceAsStream，这样就可以了！我实际上想做的是在resources文件夹中加载一个“文件夹”，循环该文件夹中的Files，并获取每个文件的Stream并读取内容……假设在运行前没有确定文件名...

问如何从jar文件中访问资源文件夹中的文件夹？
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问如何从jar文件中访问资源文件夹中的文件夹？
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