Set the path to the jar file

Sorry for the noob question, but could you tell me please what I did wrong when I try to set the path to the file in my application jar.

I have app that works with the xml file. I have the following project structure:

/project_root_directory

| / lib _

| /resources/file.XML _

| / SRC _

So, I need set my jar file, because when I correctly by running the IDE, it works well.

By default, it seems:

private File file;
private StreamResult streamResult;
file = new File("resources/file.xml");
streamResult = new StreamResult(file);

And methods where I can change the DOM structure via transformer end methods:

transformer.transform(source, streamResult);

So, I tried sets the path to the file:

private URL url;
...

file = new File(url.getPath());
streamResult = new StreamResult(file);

But it does not, because when I try to get resources by the following condition

url = getClass().getResource("resources/file.xml");

(" url = getClass().getResource("resources/file.xml "); URL = getClass () .getResource ("url = getClass().getResource("resources/file.xml "); (" resources/file.xml ');

URL - is null

OK...

I tried the following solution

InputStream input = getClass().getResourceAsStream("resources/file.xml");

Here, I myself also null...

This solution I found here

Also, I tried in absolute path

filePath = file.getAbsolutePath();
file = new File(filePath);
streamResult = new StreamResult(file);

But he also didn't work. There I got message seems to be: "Cannot find the /User/user1/Desktop/program1/resources/file.xml resource" - but it really is absoulte path to a file.

Also, I tried using System.getProperty

String filename = "file.xml";
String workingDir = System.getProperty("user.dir");
finalfile = workingDir + File.separator + "resources" + File.separator + filename;
file = new File(finalfile);

I also did test unit that finished with the 'green light', but potted his wrong with the message "unable to find /User/user1/Desktop/program1/resources/file.xm resource.

Could you tell me please what I did wrong?

file:\c:\Users\user1\Desktop\myapp.jar!\psswd.XML is what getResource() book you as the URL of the entry of the pot (notice I didn't say file, but the entrance of pot).

The meaning is: the psswd.xml of the entrance lies at the root of the C:\Users\user1\Desktop\myapp.jar pot.

Now, without taking into account my tips, you want to turn this into a file. If Java puts in front of it in the current directory. If you use Windows Explorer, you have such a file?

psswd.XML is not just a file. In any case, given that your StreamResult is supposed to write something, it may not be in the jar. You can not write to an entry in the pot. Jars is supposed to be read-only.

It's like an EXE file in writing within himself. So psswd.xml should be outside the pot, a real file and forget the things of the resource. It happened to work in the IDE, because there he was jarred not yet.

It happened to be a Windows file and a Java resource.

Tags: Java

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