blocks|key|1660760|text|不确定有没有确切的方法能给你想要的东西，但这应该会让你更接近：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1660761|import+static+org.junit.Assert.assertEquals;

import+java.net.URL;

import+org.junit.Test;

public+class+UrlTest+{

++++@Test
++++public+void+testUrl()+throws+Exception+{
++++++++URL+jarUrl+=+new+URL("jar:file:/C:/proj/parser/jar/parser.jar!/test.xml");
++++++++assertEquals("jar",+jarUrl.getProtocol());
++++++++assertEquals("file:/C:/proj/parser/jar/parser.jar!/test.xml",+jarUrl.getFile());
++++++++URL+fileUrl+=+new+URL(jarUrl.getFile());
++++++++assertEquals("file",+fileUrl.getProtocol());
++++++++assertEquals("/C:/proj/parser/jar/parser.jar!/test.xml",+fileUrl.getFile());
++++++++String[]+parts+=+fileUrl.getFile().split("!");
++++++++assertEquals("/C:/proj/parser/jar/parser.jar",+parts[0]);
++++}
}|code-block|syntax|javascript|1660762|希望这能有所帮助。|1660763|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|$]]

Not sure of any exact method that will give you what you want, but this should get you close:

<pre><code>import static org.junit.Assert.assertEquals;

import java.net.URL;

import org.junit.Test;

public class UrlTest {

 @Test
 public void testUrl() throws Exception {
 URL jarUrl = new URL("jar:file:/C:/proj/parser/jar/parser.jar!/test.xml");
 assertEquals("jar", jarUrl.getProtocol());
 assertEquals("file:/C:/proj/parser/jar/parser.jar!/test.xml", jarUrl.getFile());
 URL fileUrl = new URL(jarUrl.getFile());
 assertEquals("file", fileUrl.getProtocol());
 assertEquals("/C:/proj/parser/jar/parser.jar!/test.xml", fileUrl.getFile());
 String[] parts = fileUrl.getFile().split("!");
 assertEquals("/C:/proj/parser/jar/parser.jar", parts[0]);
 }
}
</code></pre>

Hope this helps.

blocks|key|1660816|text|如果MS-Windows没有被前导斜杠冒犯，这可能会起作用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1660817|++++final+URL+jarUrl+=
++++++++new+URL("jar:file:/C:/proj/parser/jar/parser.jar!/test.xml");
++++final+JarURLConnection+connection+=
++++++++(JarURLConnection)+jarUrl.openConnection();
++++final+URL+url+=+connection.getJarFileURL();

++++System.out.println(url.getFile());|code-block|syntax|javascript|1660818|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 might do it, if MS-Windows is not offended by a leading slash:

<pre><code> final URL jarUrl =
 new URL("jar:file:/C:/proj/parser/jar/parser.jar!/test.xml");
 final JarURLConnection connection =
 (JarURLConnection) jarUrl.openConnection();
 final URL url = connection.getJarFileURL();

 System.out.println(url.getFile());
</code></pre>

blocks|key|1660867|text|有些人可能会认为这有点“老生常谈”，但在这种情况下它会做好工作，我相信它会比在其他建议中创建所有这些对象要好得多。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1660868|String+jarUrl+=+"jar:file:/C:/proj/parser/jar/parser.jar!/test.xml";

jarUrl+=+jarUrl.substring(jarUrl.indexOf('/')%2B1,+jarUrl.indexOf('!'));|code-block|syntax|javascript|1660869|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|$]]

Some might consider this to be a bit 'hacky', but it'll do the job in that instance and i'm sure it'd perform a damn sight better than creating all those objects in the other suggestions. 

<pre><code>String jarUrl = "jar:file:/C:/proj/parser/jar/parser.jar!/test.xml";

jarUrl = jarUrl.substring(jarUrl.indexOf('/')+1, jarUrl.indexOf('!'));
</code></pre>

blocks|key|1446125|text|此解决方案将处理路径中的空格。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1446126|String+url+=+"jar:file:/C:/dir%2520with%2520spaces/myjar.jar!/resource";
String+fileUrl+=+url.substring(4,+url.indexOf('!'));
File+file+=+new+File(new+URL(fileUrl).toURI());
String+fileSystemPath+=+file.getPath();|code-block|syntax|javascript|1446127|或以URL对象开头：|1446128|...
String+fileUrl+=+url.getPath().substring(0,+url.indexOf('!'));
...|1446129|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

This solution will handle spaces in the path.

<pre><code>String url = "jar:file:/C:/dir%20with%20spaces/myjar.jar!/resource";
String fileUrl = url.substring(4, url.indexOf('!'));
File file = new File(new URL(fileUrl).toURI());
String fileSystemPath = file.getPath();
</code></pre>

or with a URL object to begin with:

<pre><code>...
String fileUrl = url.getPath().substring(0, url.indexOf('!'));
...
</code></pre>

blocks|key|1660912|text|我不得不这么做。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1660913|++++++++URL+url+=+clazz.getResource(clazz.getSimpleName()+%2B+".class");
++++++++String+proto+=+url.getProtocol();
++++++++boolean+isJar+=+proto.equals("jar");+//+see+if+it's+in+a+jar+file+URL

++++++++if(isJar)
++++++++{
++++++++++++url+=+new+URL(url.getPath());+//+this+nicely+strips+off+the+leading+jar:+
++++++++++++proto+=+url.getProtocol();
++++++++}

++++++++if(proto.equals("file"))
++++++++{
+++++++++++++if(isJar)+
++++++++++++++++//+you+can+truncate+it+at+the+last+'!'+here
++++++++}+
++++++++else+if(proto+==+"http")+{
++++++++++++...|code-block|syntax|javascript|1660914|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 just had to do this.

<pre><code> URL url = clazz.getResource(clazz.getSimpleName() + ".class");
 String proto = url.getProtocol();
 boolean isJar = proto.equals("jar"); // see if it's in a jar file URL

 if(isJar)
 {
 url = new URL(url.getPath()); // this nicely strips off the leading jar: 
 proto = url.getProtocol();
 }

 if(proto.equals("file"))
 {
 if(isJar) 
 // you can truncate it at the last '!' here
 } 
 else if(proto == "http") {
 ...
</code></pre>

blocks|key|1660923|text|++//This+code+will+work+on+both+Windows+and+Linux
++public+String+path()
++{
++URL+url1+=+getClass().getResource("");
++String+urs=url1.toString();
++urs=urs.substring(9);
++String+truepath[]=urs.split("parser.jar!");
++truepath[0]=truepath[0]%2B"parser.jar";
++truepath[0]=truepath[0].replaceAll("%2520","+");
++return+truepath[0];
++}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1660924|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code> //This code will work on both Windows and Linux
 public String path()
 {
 URL url1 = getClass().getResource("");
 String urs=url1.toString();
 urs=urs.substring(9);
 String truepath[]=urs.split("parser.jar!");
 truepath[0]=truepath[0]+"parser.jar";
 truepath[0]=truepath[0].replaceAll("%20"," ");
 return truepath[0];
 }
</code></pre>

blocks|key|1660940|text|你可以做this.This作品|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1660941|++++ClassLoader+loader+=+this.getClass().getClassLoader();
++++URL+url+=+loader.getResource("resource+name");
++++String[]+filePath+=+null;
++++String+protocol+=+url.getProtocol();
++++if(protocol.equals("jar")){
++++++++url+=+new+URL(url.getPath());
++++++++protocol+=+url.getProtocol();
++++}
++++if(protocol.equals("file")){
++++++++String[]+pathArray+=+url.getPath().split("!");
++++++++filePath+=+pathArray[0].split("/",2);
++++}
++++return+filePath[1];|code-block|syntax|javascript|1660942|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|$]]

You can do this.This works

<pre><code> ClassLoader loader = this.getClass().getClassLoader();
 URL url = loader.getResource("resource name");
 String[] filePath = null;
 String protocol = url.getProtocol();
 if(protocol.equals("jar")){
 url = new URL(url.getPath());
 protocol = url.getProtocol();
 }
 if(protocol.equals("file")){
 String[] pathArray = url.getPath().split("!");
 filePath = pathArray[0].split("/",2);
 }
 return filePath[1];
</code></pre>

I get a jar file url at runtime as:

<pre><code>jar:file:///C:/proj/parser/jar/parser.jar!/test.xml
</code></pre>

How can this be converted to a valid path as:

<pre><code>C:/proj/parser/jar/parser.jar.
</code></pre>

I have already tried using <code>File(URI)</code>, <code>getPath()</code>, <code>getFile()</code> in vain.

How do I get just the jar URL from a jar: URL containing a "!" and a specific file in the jar?

我在运行时得到一个jar文件url：jar:file:///C:/proj/parser/jar/parser.jar!/test.xml如何将其转换为有效路径，如下所示：C:/proj/parser/jar/parser.jar.我已经尝试过使用File(URI)、getPath()、getFile()，但都是徒劳的...

问如何从包含"!“的jar: URL中获取jar URL？以及jar中的特定文件？
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问如何从包含"!“的jar: URL中获取jar URL？以及jar中的特定文件？EN

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问如何从包含"!“的jar: URL中获取jar URL？以及jar中的特定文件？
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