blocks|key|1541285|text|这是故意的。“文件”的内容可能不能作为文件使用。请记住，您正在处理的类和资源可能是JAR文件或其他类型资源的一部分。类加载器不必提供资源的文件句柄，例如，jar文件可能没有扩展到文件系统中的单个文件中。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1541286|通过获取java.io.File可以完成的任何操作都可以通过将流复制到临时文件中并执行相同的操作来完成，如果java.io.File是绝对必要的话。|1541287|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

This is deliberate. The contents of the "file" may not be available as a file. Remember you are dealing with classes and resources that may be part of a JAR file or other kind of resource. The classloader does not have to provide a file handle to the resource, for example the jar file may not have been expanded into individual files in the file system.

Anything you can do by getting a java.io.File could be done by copying the stream out into a temporary file and doing the same, if a java.io.File is absolutely necessary.

blocks|key|1326258|text|如果netclient.p在JAR文件中，它将没有路径，因为该文件位于其他文件中。在这种情况下，您可以拥有的最佳路径是真正的file:/path/to/jarfile/bot.jar!/config/netclient.p。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1326259|entityMap^0|2|B|1Q|1D|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

if <code>netclient.p</code> is inside a JAR file, it won't have a path because that file is located inside other file. in that case, the best path you can have is really <code>file:/path/to/jarfile/bot.jar!/config/netclient.p</code>.

blocks|key|1541387|text|File是文件系统中文件的抽象，文件系统对JAR的内容一无所知。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1541388|尝试使用URI，我认为有一个jar://协议可能对您的目的有用。|1541389|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

A File is an abstraction for a file in a filesystem, and the filesystems don't know anything about what are the contents of a JAR.

Try with an URI, I think there's a jar:// protocol that might be useful for your purpouses.

blocks|key|1541428|text|您需要了解jar文件中的路径。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1541429|只需相对地引用它。因此，如果您有一个文件(myfile.txt)，它位于foo.jar中的\src\main\resources目录下(maven风格)。你可以这样描述它：|offset|length|style|CODE|1541430|src/main/resources/myfile.txt|code-block|syntax|javascript|1541431|如果使用jar+-tvf+myjar.jar转储jar，您将看到jar文件中的输出和相对路径，并使用正斜杠。|1541432|entityMap^0|0|19|J|0|0|4|I|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@$D|S|E|T|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|U|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|V|8|@$D|W|E|X|F|G]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|P|$]]

You need to understand the path within the jar file. 
Simply refer to it relative. So if you have a file (myfile.txt), located in foo.jar under the <code>\src\main\resources</code> directory (maven style). You would refer to it like:

<pre><code>src/main/resources/myfile.txt
</code></pre>

If you dump your jar using <code>jar -tvf myjar.jar</code> you will see the output and the relative path within the jar file, and use the FORWARD SLASHES.

blocks|key|1541478|text|我花了一段时间解决这个问题，因为奇怪的是，我找不到真正有效的解决方案！工作目录通常不是JAR的目录，特别是当从Windows下的开始菜单运行JAR+(或任何程序)时。这就是我所做的，它对从JAR外部运行的.class文件的效果与对JAR的效果一样好。(我只在Windows7下进行了测试。)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1541479|try+{
++++//Attempt+to+get+the+path+of+the+actual+JAR+file,+because+the+working+directory+is+frequently+not+where+the+file+is.
++++//Example:+file:/D:/all/Java/TitanWaterworks/TitanWaterworks-en.jar!/TitanWaterworks.class
++++//Another+example:+/D:/all/Java/TitanWaterworks/TitanWaterworks.class
++++PROGRAM_DIRECTORY+=+getClass().getClassLoader().getResource("TitanWaterworks.class").getPath();+//+Gets+the+path+of+the+class+or+jar.

++++//Find+the+last+!+and+cut+it+off+at+that+location.+If+this+isn't+being+run+from+a+jar,+there+is+no+!,+so+it'll+cause+an+exception,+which+is+fine.
++++try+{
++++++++PROGRAM_DIRECTORY+=+PROGRAM_DIRECTORY.substring(0,+PROGRAM_DIRECTORY.lastIndexOf('!'));
++++}+catch+(Exception+e)+{+}

++++//Find+the+last+/+and+cut+it+off+at+that+location.
++++PROGRAM_DIRECTORY+=+PROGRAM_DIRECTORY.substring(0,+PROGRAM_DIRECTORY.lastIndexOf('/')+%2B+1);
++++//If+it+starts+with+/,+cut+it+off.
++++if+(PROGRAM_DIRECTORY.startsWith("/"))+PROGRAM_DIRECTORY+=+PROGRAM_DIRECTORY.substring(1,+PROGRAM_DIRECTORY.length());
++++//If+it+starts+with+file:/,+cut+that+off,+too.
++++if+(PROGRAM_DIRECTORY.startsWith("file:/"))+PROGRAM_DIRECTORY+=+PROGRAM_DIRECTORY.substring(6,+PROGRAM_DIRECTORY.length());
}+catch+(Exception+e)+{
++++PROGRAM_DIRECTORY+=+"";+//Current+working+directory+instead.
}|code-block|syntax|javascript|1541480|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|$]]

I spent a while messing around with this problem, because no solution I found actually worked, strangely enough! The working directory is frequently not the directory of the JAR, especially if a JAR (or any program, for that matter) is run from the Start Menu under Windows. So here is what I did, and it works for .class files run from outside a JAR just as well as it works for a JAR. (I only tested it under Windows 7.)

<pre><code>try {
 //Attempt to get the path of the actual JAR file, because the working directory is frequently not where the file is.
 //Example: file:/D:/all/Java/TitanWaterworks/TitanWaterworks-en.jar!/TitanWaterworks.class
 //Another example: /D:/all/Java/TitanWaterworks/TitanWaterworks.class
 PROGRAM_DIRECTORY = getClass().getClassLoader().getResource("TitanWaterworks.class").getPath(); // Gets the path of the class or jar.

 //Find the last ! and cut it off at that location. If this isn't being run from a jar, there is no !, so it'll cause an exception, which is fine.
 try {
 PROGRAM_DIRECTORY = PROGRAM_DIRECTORY.substring(0, PROGRAM_DIRECTORY.lastIndexOf('!'));
 } catch (Exception e) { }

 //Find the last / and cut it off at that location.
 PROGRAM_DIRECTORY = PROGRAM_DIRECTORY.substring(0, PROGRAM_DIRECTORY.lastIndexOf('/') + 1);
 //If it starts with /, cut it off.
 if (PROGRAM_DIRECTORY.startsWith("/")) PROGRAM_DIRECTORY = PROGRAM_DIRECTORY.substring(1, PROGRAM_DIRECTORY.length());
 //If it starts with file:/, cut that off, too.
 if (PROGRAM_DIRECTORY.startsWith("file:/")) PROGRAM_DIRECTORY = PROGRAM_DIRECTORY.substring(6, PROGRAM_DIRECTORY.length());
} catch (Exception e) {
 PROGRAM_DIRECTORY = ""; //Current working directory instead.
}
</code></pre>

blocks|key|1541676|text|下面的路径适用于我：classpath:/path/to/resource/in/jar|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1541677|entityMap^0|A|Y|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|$]]

The following path worked for me: <code>classpath:/path/to/resource/in/jar</code>

blocks|key|1326618|text|加载资源时，请注意以下各项之间的区别：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1326619|getClass().getClassLoader().getResource("com/myorg/foo.jpg")+//relative+path|code-block|syntax|javascript|1326620|和|1326621|getClass().getResource("/com/myorg/foo.jpg"));+//note+the+slash+at+the+beginning|1326622|我想，这种混乱导致了加载资源时出现的大多数问题。|1326623|1326624|此外，当您加载图像时，使用getResourceAsStream()会更容易|offset|length|style|CODE|1326625|BufferedImage+image+=+ImageIO.read(getClass().getResourceAsStream("/com/myorg/foo.jpg"));|1326626|当您确实需要从JAR归档文件中加载(非图像)文件时，您可以尝试这样做：|1326627|File+file+=+null;
String+resource+=+"/com/myorg/foo.xml";
URL+res+=+getClass().getResource(resource);
if+(res.getProtocol().equals("jar"))+{
++++try+{
++++++++InputStream+input+=+getClass().getResourceAsStream(resource);
++++++++file+=+File.createTempFile("tempfile",+".tmp");
++++++++OutputStream+out+=+new+FileOutputStream(file);
++++++++int+read;
++++++++byte[]+bytes+=+new+byte[1024];

++++++++while+((read+=+input.read(bytes))+!=+-1)+{
++++++++++++out.write(bytes,+0,+read);
++++++++}
++++++++out.close();
++++++++file.deleteOnExit();
++++}+catch+(IOException+ex)+{
++++++++Exceptions.printStackTrace(ex);
++++}
}+else+{
++++//this+will+probably+work+in+your+IDE,+but+not+from+a+JAR
++++file+=+new+File(res.getFile());
}

if+(file+!=+null+&&+!file.exists())+{
++++throw+new+RuntimeException("Error:+File+"+%2B+file+%2B+"+not+found!");
}|1326628|entityMap^0|0|0|0|0|0|0|D|L|0|0|0|0^^$0|@$1|2|3|4|5|6|7|11|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|12|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|13|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|14|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|15|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|16|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|17|8|@$P|18|Q|19|R|S]]|9|@]|A|$]]|$1|T|3|U|5|D|7|1A|8|@]|9|@]|A|$E|F]]|$1|V|3|W|5|6|7|1B|8|@]|9|@]|A|$]]|$1|X|3|Y|5|D|7|1C|8|@]|9|@]|A|$E|F]]|$1|Z|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]]|10|$]]

When loading a resource make sure you notice the difference between:
<pre><code>getClass().getClassLoader().getResource(&quot;com/myorg/foo.jpg&quot;) //relative path
</code></pre>
and
<pre><code>getClass().getResource(&quot;/com/myorg/foo.jpg&quot;)); //note the slash at the beginning
</code></pre>
I guess, this confusion is causing most of problems when loading a resource.
<hr />
Also, when you're loading an image it's easier to use <code>getResourceAsStream()</code>:
<pre><code>BufferedImage image = ImageIO.read(getClass().getResourceAsStream(&quot;/com/myorg/foo.jpg&quot;));
</code></pre>
When you really have to load a (non-image) file from a JAR archive, you might try this:
<pre><code>File file = null;
String resource = &quot;/com/myorg/foo.xml&quot;;
URL res = getClass().getResource(resource);
if (res.getProtocol().equals(&quot;jar&quot;)) {
 try {
 InputStream input = getClass().getResourceAsStream(resource);
 file = File.createTempFile(&quot;tempfile&quot;, &quot;.tmp&quot;);
 OutputStream out = new FileOutputStream(file);
 int read;
 byte[] bytes = new byte[1024];

 while ((read = input.read(bytes)) != -1) {
 out.write(bytes, 0, read);
 }
 out.close();
 file.deleteOnExit();
 } catch (IOException ex) {
 Exceptions.printStackTrace(ex);
 }
} else {
 //this will probably work in your IDE, but not from a JAR
 file = new File(res.getFile());
}

if (file != null &amp;&amp; !file.exists()) {
 throw new RuntimeException(&quot;Error: File &quot; + file + &quot; not found!&quot;);
}
</code></pre>

blocks|key|1541806|text|在jar文件中，资源绝对位于包层次结构(而不是文件系统层次结构)中。因此，如果您让类com.example.Sweet加载名为./default.conf的资源，那么该资源的名称将指定为/com/example/default.conf。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1541807|但如果它在罐子里那它就不是文件..。|1541808|entityMap^0|1|3|16|H|1R|E|2L|P|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]|$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@]|D|@]|E|$]]|$1|H|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|I|$]]

When in a <code>jar</code> file, the resource is located absolutely in the package hierarchy (not file system hierarchy). So if you have class <code>com.example.Sweet</code> loading a resource named <code>./default.conf</code> then the resource's name is specified as <code>/com/example/default.conf</code>.
But if it's in a jar then it's not a File ...

blocks|key|1541857|text|答案只有一行：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1541858|String+path+=+this.getClass().getClassLoader().getResource(<resourceFileName>).toExternalForm()|code-block|syntax|javascript|1541859|基本上，getResource方法给出了URL。您可以通过调用toExternalForm()从这个URL中提取路径。|offset|length|style|CODE|1541860|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/8/docs/api/java/lang/Class.html#getResource-java.lang.String-|1|https://docs.oracle.com/javase/8/docs/api/java/net/URL.html#toExternalForm--^0|0|0|4|B|V|G|4|B|0|V|G|1|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@$I|Z|J|10|K|L]|$I|11|J|12|K|L]]|9|@$I|13|J|14|1|15]|$I|16|J|17|1|18]]|A|$]]|$1|M|3|-4|5|6|7|19|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]|U|$5|P|Q|R|A|$S|V]]]]

The one line answer is -
<pre><code>String path = this.getClass().getClassLoader().getResource(&lt;resourceFileName&gt;).toExternalForm()
</code></pre>
Basically <a href="https://docs.oracle.com/javase/8/docs/api/java/lang/Class.html#getResource-java.lang.String-" rel="nofollow noreferrer"><code>getResource</code></a> method gives the URL. From this URL you can extract the path by calling <a href="https://docs.oracle.com/javase/8/docs/api/java/net/URL.html#toExternalForm--" rel="nofollow noreferrer"><code>toExternalForm()</code></a>.

blocks|key|1541937|text|这是来自用户Tombart的相同代码，带有流刷新和关闭，以避免从jar资源中复制不完整的临时文件内容，并具有唯一的临时文件名。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1541938|File+file+=+null;
String+resource+=+"/view/Trial_main.html"+;
URL+res+=+getClass().getResource(resource);
if+(res.toString().startsWith("jar:"))+{
++++try+{
++++++++InputStream+input+=+getClass().getResourceAsStream(resource);
++++++++file+=+File.createTempFile(new+Date().getTime()%2B"",+".html");
++++++++OutputStream+out+=+new+FileOutputStream(file);
++++++++int+read;
++++++++byte[]+bytes+=+new+byte[1024];

++++++++while+((read+=+input.read(bytes))+!=+-1)+{
++++++++++++out.write(bytes,+0,+read);
++++++++}
++++++++out.flush();
++++++++out.close();
++++++++input.close();
++++++++file.deleteOnExit();
++++}+catch+(IOException+ex)+{
++++++++ex.printStackTrace();
++++}
}+else+{
++++//this+will+probably+work+in+your+IDE,+but+not+from+a+JAR
++++file+=+new+File(res.getFile());
}|code-block|syntax|javascript|1541939|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|$]]

This is same code from user Tombart with stream flush and close to avoid incomplete temporary file content copy from jar resource and to have unique temp file names.
<pre><code>File file = null;
String resource = &quot;/view/Trial_main.html&quot; ;
URL res = getClass().getResource(resource);
if (res.toString().startsWith(&quot;jar:&quot;)) {
 try {
 InputStream input = getClass().getResourceAsStream(resource);
 file = File.createTempFile(new Date().getTime()+&quot;&quot;, &quot;.html&quot;);
 OutputStream out = new FileOutputStream(file);
 int read;
 byte[] bytes = new byte[1024];

 while ((read = input.read(bytes)) != -1) {
 out.write(bytes, 0, read);
 }
 out.flush();
 out.close();
 input.close();
 file.deleteOnExit();
 } catch (IOException ex) {
 ex.printStackTrace();
 }
} else {
 //this will probably work in your IDE, but not from a JAR
 file = new File(res.getFile());
}
 
</code></pre>

blocks|key|1541982|text|可能有点晚了，但您可以使用我的库KResourceLoader从您的jar中获取资源：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1541983|File+resource+=+getResource("file.txt")|code-block|syntax|javascript|1541984|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/lamba92/KResourceLoader^0|G|F|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

It may be a little late but you may use my library <a href="https://github.com/lamba92/KResourceLoader" rel="nofollow noreferrer">KResourceLoader</a> to get a resource from your jar:

<pre><code>File resource = getResource("file.txt")
</code></pre>

blocks|key|1326831|text|也许这种方法可以用来快速求解。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1326832|public+class+TestUtility
{+
++++public+static+File+getInternalResource(String+relativePath)
++++{
++++++++File+resourceFile+=+null;
++++++++URL+location+=+TestUtility.class.getProtectionDomain().getCodeSource().getLocation();
++++++++String+codeLocation+=+location.toString();
++++++++try{
++++++++++++if+(codeLocation.endsWith(".jar"){
++++++++++++++++//Call+from+jar
++++++++++++++++Path+path+=+Paths.get(location.toURI()).resolve("../classes/"+%2B+relativePath).normalize();
++++++++++++++++resourceFile+=+path.toFile();
++++++++++++}else{
++++++++++++++++//Call+from+IDE
++++++++++++++++resourceFile+=+new+File(TestUtility.class.getClassLoader().getResource(relativePath).getPath());
++++++++++++}
++++++++}catch(URISyntaxException+ex){
++++++++++++ex.printStackTrace();
++++++++}
++++++++return+resourceFile;
++++}
}|code-block|syntax|javascript|1326833|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|$]]

Maybe this method can be used for quick solution.

<pre><code>public class TestUtility
{ 
 public static File getInternalResource(String relativePath)
 {
 File resourceFile = null;
 URL location = TestUtility.class.getProtectionDomain().getCodeSource().getLocation();
 String codeLocation = location.toString();
 try{
 if (codeLocation.endsWith(".jar"){
 //Call from jar
 Path path = Paths.get(location.toURI()).resolve("../classes/" + relativePath).normalize();
 resourceFile = path.toFile();
 }else{
 //Call from IDE
 resourceFile = new File(TestUtility.class.getClassLoader().getResource(relativePath).getPath());
 }
 }catch(URISyntaxException ex){
 ex.printStackTrace();
 }
 return resourceFile;
 }
}
</code></pre>

blocks|key|1542041|text|遵循代码！|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1542042|/src/main/resources/file|1542043|streamToFile(getClass().getClassLoader().getResourceAsStream("file"))

public+static+File+streamToFile(InputStream+in)+{
++++if+(in+==+null)+{
++++++++return+null;
++++}

++++try+{
++++++++File+f+=+File.createTempFile(String.valueOf(in.hashCode()),+".tmp");
++++++++f.deleteOnExit();

++++++++FileOutputStream+out+=+new+FileOutputStream(f);
++++++++byte[]+buffer+=+new+byte[1024];

++++++++int+bytesRead;
++++++++while+((bytesRead+=+in.read(buffer))+!=+-1)+{
++++++++++++out.write(buffer,+0,+bytesRead);
++++++++}

++++++++return+f;
++++}+catch+(IOException+e)+{
++++++++LOGGER.error(e.getMessage(),+e);
++++++++return+null;
++++}
}|code-block|syntax|javascript|1542044|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

follow code!
<h1>/src/main/resources/file</h1>
<pre class="lang-java prettyprint-override"><code>streamToFile(getClass().getClassLoader().getResourceAsStream(&quot;file&quot;))

public static File streamToFile(InputStream in) {
 if (in == null) {
 return null;
 }

 try {
 File f = File.createTempFile(String.valueOf(in.hashCode()), &quot;.tmp&quot;);
 f.deleteOnExit();

 FileOutputStream out = new FileOutputStream(f);
 byte[] buffer = new byte[1024];

 int bytesRead;
 while ((bytesRead = in.read(buffer)) != -1) {
 out.write(buffer, 0, bytesRead);
 }

 return f;
 } catch (IOException e) {
 LOGGER.error(e.getMessage(), e);
 return null;
 }
}
</code></pre>

blocks|key|1326899|text|这个类函数可以根据.jar文件中的相对文件路径得到文件的绝对路径。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1326900|public+class+Utility+{

++++public+static+void+main(String[]+args)+throws+Exception+{
++++++++Utility+utility+=+new+Utility();
++++++++String+absolutePath+=+utility.getAbsolutePath("./absolute/path/to/file");
++++}

++++public+String+getAbsolutePath(String+relativeFilePath)+throws+IOException+{
++++++++URL+url+=+this.getClass().getResource(relativeFilePath);
++++++++return+url.getPath();
++++}
}|code-block|syntax|javascript|1326901|如果目标文件与Utility.java位于同一目录，则您的输入将为./file.txt。|offset|length|style|CODE|1326902|当您在getAbsolutePath()上只输入/时，它返回/Users/user/PROJECT_NAME/target/classes/。这意味着您可以像这样选择文件的/com/example/file.txt。|1326903|entityMap^0|0|0|7|C|X|A|0|O|1|U|14|2E|L|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]|$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|X|8|@$I|Y|J|Z|K|L]|$I|10|J|11|K|L]|$I|12|J|13|K|L]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|P|$]]

This Class function can get absolute file path by relative file path in .jar file.
<pre><code>
public class Utility {

 public static void main(String[] args) throws Exception {
 Utility utility = new Utility();
 String absolutePath = utility.getAbsolutePath(&quot;./absolute/path/to/file&quot;);
 }

 public String getAbsolutePath(String relativeFilePath) throws IOException {
 URL url = this.getClass().getResource(relativeFilePath);
 return url.getPath();
 }
}

</code></pre>
If target file is same directory with <code>Utility.java</code>, your input will be <code>./file.txt</code>.
When you input only <code>/</code> on getAbsolutePath(), it returns <code>/Users/user/PROJECT_NAME/target/classes/</code>. This means you can select the file like this <code>/com/example/file.txt</code>.

I am trying to get a path to a Resource but I have had no luck. 

This works (both in IDE and with the JAR) but this way I can't get a path to a file, only the file contents: 

<pre><code>ClassLoader classLoader = getClass().getClassLoader();
PrintInputStream(classLoader.getResourceAsStream("config/netclient.p"));
</code></pre>

If I do this:

<pre><code>ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("config/netclient.p").getFile());
</code></pre>

The result is: 
<code>java.io.FileNotFoundException: file:/path/to/jarfile/bot.jar!/config/netclient.p (No such file or directory)</code>

Is there a way to get a path to a resource file?

How to get a path to a resource in a Java JAR file

文件系统

Windows

Java

我正在尝试获取一个资源的路径，但我没有运气。这是有效的(在IDE和JAR中)，但是通过这种方式，我不能获得文件的路径，只能获得文件内容：ClassLoader classLoader = getClass().getClassLoader();PrintInputStream(classLoader.getResour...

问如何获取Java JAR文件中资源的路径
EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何获取Java JAR文件中资源的路径EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何获取Java JAR文件中资源的路径
EN