Get filepath in web application

Try this, it gives you the path which you are executing your program -

public class GetExecutionPath
{
    public static void main(String args[])
    {
	try
	{
	    String executionPath = System.getProperty("user.dir");
	    System.out.print(executionPath);
	} catch (Exception e)
	{
	    
	}
    }
}

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I've already tried it.
I can get the file path if my program is stand alone.
For web application, I just got C:\Eclipse, since user.dir contains the current working directory (ie. where the vm was run from).

But you want the location of the xml?

It is like I want to retrieve xml from my java bean, so, I want to get that path to my java bean. In my standard program. I used this:

static String curDir = System.getProperty("user.dir");
static String resources = ("\\WebContent\\WEB-INF\\classes\\resources");
and I give the path for my xml file like this
File file = new File(curDir + resources + "myxml.xml") and it works.

But for my web application, System.getProperty("user.dir") only returns C:\Eclipse where VM runs.

So, I want to know how to get file path in web application.

http://helpdesk.objects.com.au/java/how-to-get-the-path-of-a-file-in-a-web-application

Thanks.
But my JavaBean program can't extend to Servletcontext.

you'll need to pass it a context (or request) to determine the path

>But my JavaBean program can't extend to Servletcontext.
You are right. But you could create a load-on-startup servlet. In its init method  you could set the ServletContext into a field of a static utility class.  Your javabean could access that.  
 

Thanks.

Can you explain me a bit more about load-on-startup servlet. How can I do this?