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

Question

I get a jar file url at runtime as:

jar:file:///C:/proj/parser/jar/parser.jar!/test.xml

How can this be converted to a valid path as:

C:/proj/parser/jar/parser.jar.

I have already tried using File(URI), getPath(), getFile() in vain.

score 41 · Answer 1 · answered Dec 31 '08 at 11:22

41

This might do it, if MS-Windows is not offended by a leading slash:

    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());

answered Dec 31 '08 at 11:22

starblue

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I am going with exactly this solution, although I wish there was a way to do this without opening the jar (or parsing directly, of course); seems like URL needs to understand this weird ! syntax, or else a subclass needs to exist which understands it. – skiphoppy Jul 15 '09 at 17:23
6

(In Java7) JarURLConnection doesn't open the JAR file when you execute the above code. It just parses the URL on your behalf. – Gili Nov 04 '11 at 18:36
@Marcono1234 This will throw an exception if the URI points to something inside a jar, thus having exclamation marks , as is the case for OP. – Dmitry Avtonomov May 14 '19 at 16:29
@DmitryAvtonomov, you are right, my suggestion to use `Paths.get` will not work. I removed that comment. – Marcono1234 May 18 '19 at 10:31

score 3 · Answer 2 · answered Dec 31 '08 at 11:09

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

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]);
    }
}

Hope this helps.

Very close, but the usage of split() is too much low-level parsing for me. Sun's URL implementation doesn't seem to provide methods to deal with this weird jarfile.jar!path/to/specific/file syntax for Sun's weird jar URLs; the whole shebang (pun intended) is returned by getPath(). But JarURLConnection, as starblue mentioned, seems to work, although only through opening up the jar. — skiphoppy, Jul 15 '09 at 17:22

score 1 · Answer 3 · answered Dec 31 '08 at 11:35

1

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.

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

jarUrl = jarUrl.substring(jarUrl.indexOf('/')+1, jarUrl.indexOf('!'));

answered Dec 31 '08 at 11:35

James Camfield

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This will fail miserably if there are any special (percent encoded) characters in the URL. – cbley Jun 21 '16 at 09:34

dcstraw · Answer 4 · 2009-05-21T15:20:33.323

1

This solution will handle spaces in the path.

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();

or with a URL object to begin with:

...
String fileUrl = url.getPath().substring(0, url.indexOf('!'));
...

edited May 21 '09 at 15:20

answered May 21 '09 at 15:14

dcstraw

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score 1 · Answer 5 · answered Feb 08 '12 at 07:00

I just had to do this.

        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") {
            ...

score 0 · Answer 6 · answered Aug 21 '12 at 07:36

  //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];
  }

Hari Chandu Vakacharla · Answer 7 · 2015-09-14T04:09:04.833

-1

You can do this.This works

    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];

edited Sep 14 '15 at 04:09

answered Sep 14 '15 at 03:41

Hari Chandu Vakacharla

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Not working: the string part after ! is still not removed – tribbloid Sep 03 '16 at 22:56

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

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