Reading File In JAR using Relative Path

Question

I have some text configuration file that need to be read by my program. My current code is:

protected File getConfigFile() {
    URL url = getClass().getResource("wof.txt");
    return new File(url.getFile().replaceAll("%20", " "));
}

This works when I run it locally in eclipse, though I did have to do that hack to deal with the space in the path name. The config file is in the same package as the method above. However, when I export the application as a jar I am having problems with it. The jar exists on a shared, mapped network drive Z:. When I run the application from command line I get this error:

java.io.FileNotFoundException: file:\Z:\apps\jar\apps.jar!\vp\fsm\configs\wof.txt

How can I get this working? I just want to tell java to read a file in the same directory as the current class.

Thanks, Jonah

Bahrain

Answer 1

When the file is inside a jar, you can't use the File class to represent it, since it is a jar: URI. Instead, the URL class itself already gives you with openStream() the possibility to read the contents.

Or you can shortcut this by using getResourceAsStream() instead of getResource().

To get a BufferedReader (which is easier to use, as it has a readLine() method), use the usual stream-wrapping:

InputStream configStream = getClass().getResourceAsStream("wof.txt");
BufferedReader configReader = new BufferedReader(new InputStreamReader(configStream, "UTF-8"));

Instead of "UTF-8" use the encoding actually used by the file (i.e. which you used in the editor).

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Another point: Even if you only have file: URIs, you should not do the URL to File-conversion yourself, instead use new File(url.toURI()). This works for other problematic characters as well.

Operatic

Paulo, thanks. The problem is I ultimately need to get a BufferedReader. How can I do that if I use getResourceAsStream()? – Bahrain

new BufferedReader(new InputStreamReader(getClass().getResourceAsStream("wof.txt"))) – Crispa