Open an XML file in tomcat with a NOT servlet class

Hello,
I have an xml file placed in /my_app/WEB-INF/classes/data/cfg.xml and i would like to open it with the XMLReader java class placed in /my_app/WEB-INF/classes/utility/XMLReader.java. To open the file i do this:

The "doc" object rest null... Can you tell me how open an XML file in Tomcat with a not servlet class?
Thanks in advantage

Nothing Tomcat-specific here, so moved to the Servlets forum.

Your path and the location of the file within the classpath do not match.

The directory "/my_app/WEB-INF/classes" is in your classpath. To load that file from the classpath -- which is what getResourceAsStream does -- you would have to specify its path relative to the classpath. In your example that would be "/data/cfg.xml".

I've tried also to like this but the resulting object is always null... this means that the file is not found isn't it?

Maybe my classpath is .../Apache Software Foundation/Tomcat 6.0/bin/ cause when i tried to open the XML file creating a File object that was the path in wich Tomcat looked (without find nothing)

Maybe my classpath is .../Apache Software Foundation/Tomcat 6.0/bin/ cause when i tried to open the XML file creating a File object that was the path in wich Tomcat looked (without find nothing)

That is a file path, not a classpath; those concepts are not related. (As an aside, in a web app you always need to use absolute paths for file I/O, because there is no default directory to use.)

My web app should be installed on different machines...use absolute path is quite difficult...

Ulf Dittmer wrote:

As an aside, in a web app you always need to use absolute paths for file I/O, because there is no default directory to use

If there is no default directory and I use an absolute path to a file, in different installations my web application will not work

My web app should be installed on different machines...use absolute path is quite difficult...

I was referring to what you wrote earlier: "Maybe my classpath is .../Apache Software Foundation/Tomcat 6.0/bin/ cause when i tried to open the XML file creating a File object that was the path in wich Tomcat looked" which led me to believe that you were using relative paths in a web app, which doesn't work due to the lack of a default directory.

Absolute paths should not be completely fixed - part of them should be the context path. That way it doesn't matter what the directory is named in which the web app is installed - you'll always get the correct one.

Ok but how can i do this? I've tried the method above without results (the document object is always null)

hehe

Document doc = builder.parse(XMLReader.class.getResourceAsStream("cfg.xml"));

try this...

dunno if it makes any difference but this is how I usually do it.

Sean

Thank you very much, it seems that like this:

the Document object isn't null, in fact the "Opened" message is printed... anyway the class can't find anything in the document... and if i do:

I get this message: "Document: #null"

I get this message: "Document: #null"

This does not mean that the doc is null. That is just what the default implementtation of doc.toString() will print out.

If you want to see the actual xml then you will need to serialise it.

Sean

Thank you very much with your method all works perfectly! Not i can open XLM files in Tomcat and not ^_^

One more thing....

There is a lovely table in the JavaDocs for the org.w3c.dom.Node interface that shows what each kind of Node toString() operation will display plus a bunch of very useful information. The different kinds of Node behave quite differently.

Getting very familiar with that table will help you resolve similar problems in your continued XML adventure.

Bill