Hi, 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...
views:
2287answers:
5
+2
A:
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.
toolkit
2008-12-31 11:09:06
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
2009-07-15 17:22:42
Because I know diddly-squat about the format of this bizarre jar:file:// URL I am dealing with, which Sun invented and doesn't appear to have documented, and so therefore I can't possibly know if my parsing will work in all circumstances. At least if I use Sun's Java URL libraries, I have a slight chance that it'll work for more cases I might encounter. Maybe. If the libraries are good. They'll be more widely used and tested than anything I write, that's for sure.
skiphoppy
2009-07-15 16:03:57
I voted you up because this is an important question, even though it is definitely a bad suggestion.
skiphoppy
2009-07-15 17:24:25
There are other ways to raise important questions, aside from telling one to "continue with his life"
Leonel
2009-07-15 17:53:59
+2
A:
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());
starblue
2008-12-31 11:22:04
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
2009-07-15 17:23:52
+1
A:
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('!'));
James Camfield
2008-12-31 11:35:14
A:
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('!'));
...
dcstraw
2009-05-21 15:14:04