Help With Java and Reading An XML File

dk345

dk345

dk345

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Hi guys,

I've recently inherited a Java application that needs to be deployed on the web using Java Web Start. As it stands, the program must be run from a directory that has a Data folder along with the .jar file. This is because the application needs to access the resources stored in the data folder.

Now, when I try and turn this in to a Web Start App, it doesn't like having to read from the external folder. As it stands, the code to read in the XML is as follows:

Document doc = docBuilder.parse(new File("data" + File.separator + "configFile1.xml"));

So, I'm trying to add everything that is needed to the Jar file, so that when Web Start executes the Jar file, everything it needs is in there. So, if I move configFile1.xml to the root of my jar file, I assumed I could just alter the code to this:

Document doc = docBuilder.parse(new File("/configFile1.xml"));

And it would look for the file in the root of the Jar, where it is. But when I try this, I get a Exception telling me that the file couldn't be found.

So now I am stuck, and I've bern looking at it on and off for about a week now, it's hurting my head. I'd rather not re-write how it parses the XML files, so I'd like to know if this is do-able how I'm trying, or am I completely off?

A Million Internet Cookies to anyone who can help me...
 
The files aren't relative to the jar file, but relative to your filesystem, which the .jar file just extends. So "/something" is looking at the root of your disk for the file where the .jar resides.

If you've placed the file 3 subdirectories up from where the code is inside the jar/package structure, then use relative pathing from your code to find it like this "../../../configFile1.xml" as that would always be constant wherever the .jar file lived.
 
topdog@OC said:
The files aren't relative to the jar file, but relative to your filesystem, which the .jar file just extends. So "/something" is looking at the root of your disk for the file where the .jar resides.

If you've placed the file 3 subdirectories up from where the code is inside the jar/package structure, then use relative pathing from your code to find it like this "../../../configFile1.xml" as that would always be constant wherever the .jar file lived.
Click to expand...

Thanks for your help topdog, I understand what you're saying. I've made the changes but I'm still getting an error, so I've probably done something wrong.

The Class that needs the XML file is here:

Root --> Entities --> Factory --> Class.java

And the XML files are in Root. So I changed my code to:

Document doc = docBuilder.parse(new File("../../../configFile1.xml"));

But it still says that the file couldn't be found :(
 
Is it a resource ?

Are you embedding the XML file inside the jar ? If so, then treating it as a resource is the way to go. Start from here: http://docs.oracle.com/javase/1.5.0/docs/guide/lang/resources.html

Once you've read the XML, walking the tree is your next fun. Do you have an XSD of the XML schema ? If so, look at JAXB which will generate a set of java bean classes to walk the tree instead of the fugly document and nodelist classes.
 
Oh, blah, contents in .jar files aren't able to use the standard File objects, you need to read them in using getResourceAsStream() instead (a method of ClassLoader). Actually then, the "/configFile1.xml" approach would work if the file is at the beginning or root of the classpath, and won't be looking at the beginning of the disk.
 
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