Read File from jar file

[Eddie Howard]

Hi

I am creating some XML in my code....... I gave an xsd that I want to validated my XML against. I have shipped xsd as part of my jar file but I am having trouble accessing xsd from jar....

Can anyone pls advise..

My XSL/XSK code looks like....

DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
factory.setNamespaceAware(true);
factory.setValidating(true);
factory.setAttribute(
"http://java.sun.com/xml/jaxp/properties/schemaLanguage",
"http://www.w3.org/2001/XMLSchema");
factory.setAttribute(
"http://java.sun.com/xml/jaxp/properties/schemaSource",
new File(new URI("com.package.schemFile.xsd")));
DocumentBuilder builder = factory.newDocumentBuilder();
builder.setErrorHandler(new GenericErrorHandler());
builder.parse(new InputSource(new StringReader(xmlString)));

How ever I am getting an error reading URI\xsd file.... can anyone advise

Rgds
Ed

[Peter Chase]

I think you probably want ClassLoader.getResourceAsStream()

[Rahul Bhattacharjee]

If you have put your XSD in package com.package with name schemFile.xsd and assume that java runtime would identify this , then I do not think this would work.Runtime would look for classes in this way not for other resources.

You might try putting the XSD file in the root of the jar and changing the file as below.

factory.setAttribute(
"http://java.sun.com/xml/jaxp/properties/schemaSource",
new File(schemFile.xsd"));

[ August 13, 2007: Message edited by: Rahul Bhattacharjee ]

[saurabh suman]

Hi Eddie,
I am not sure what exact error you are getting. if you are getting acess denied then you need to give permission of read to the jar file.
If possible can you post your error , that will help me more

[Rahul Bhattacharjee]

If you are not running the program with security manager enabled , then no problem.You will not get access controller exception.
Have you tried the solution that I suggested you in my previous post?

[Eddie Howard]

Hi Guys

Thank for comments. Original details were a bit sketchy so for that I apologies.
My original code didn't work as I could find xsd file.

I tried a few approaches so let me try yo explain

EFFORT 1
I moved it up to root of jar file and used following code:

URL url = ClassLoader.getSystemResource("gsk2c.xsd") ;
.....
factory.setAttribute(
                                                      "http://java.sun.com/xml/jaxp/properties/schemaSource",
                                                      new File(uri));

So basically using class loader to get access to resource, creating URI and passing URI to new file.
I work fine in my IDE when I am developing (gsk2c.xsd is local file) but when I test shipped jar, I get following error.
Exception : com.gsk.gskgaa.exceptions.GSKGAAException: java.lang.IllegalArgument
Exception: URI is not hierarchical

From some research I gather this is widely reported. And of course there is problem because I am using new File(uri), but when deployed my uri NOT a file.... it is embedded in jar file...

Now I as use CLassLoader.getResourceAsINputStream, this give me accurate handle to xsd...... however when setting attribute, I require either a file or file location (as far as I know...), I tried reading actually xsd contents and passin it to setAttribte but I got error again there.... cannot read xsd file etc

EFFORT2
I hoped to explicity read file using about approach, so i took file out of jar & added to classpath, however clasloader still looking for file in jar.... I get following NPE as classloader trying to read file from
URL jar:file:/C:/NewAdapter/lib/gskgaa3.012Beta.jar!/gsk2c.xsd

FINALLY EFFORT3
I have an external properties file. I am left with specifying explicitly where file is & using the file location when setting attribute. This is very crude but all I'm left with....

So to sumarise.....
I can create an inputstream from file in jar..... but I don't where to go then..... as far as I know I need either a File object or a file location....

Any comments

[David O'Meara]

You did skip one possibility:

you are able to pass a URI when it is on the local file system, but you missed the fact that gskgaa3.012Beta.jar!/gsk2c.xsd represents the URL for the file within the JAR.

[Eddie Howard]

Ok.....
I will try that.....
Thanks!!

[Eddie Howard]

Ok I have a URI from which I can create a URL......

URL url = ClassLoader.getSystemResource("gsk2c.xsd") ; 
URI uri = new URI(url.toString()); 

However when I use:

factory.setAttribute("http://java.sun.com/xml/jaxp/properties/schemaSource",
 new File(uri));

I get URI is not hierarchical at this line. However when I use

factory.setAttribute(
    "http://java.sun.com/xml/jaxp/properties/schemaSource",
     new File(uri.toString()));

I get my old favourite..... schema_reference.4: Failed to read schema document 'jar:file:\C:\Projects\GSK\GSKGAAAdapter\lib\gsk2c.jar!\gsk2c.xsd', because 1) could not find the document; 2) the document could not be read; 3) the root element of the document is not <xsd:schema>.

I am assumeing

it's not 1) as it knows where doc is

its not 3) as .xsd has worked fine for me in past

it could be 2) ???

Have I covered your approach or am I still missing something?

[David O'Meara]

You cannot get a File reference to a File in a Jar, it just won't work. You can, however get a URL or an InputStream. These are your only real options to try if you want the file in your jar.

[Nirav Naik]

The idea is that if the class (the code) was able to be loaded from the jar file, then you ought to be able to ask that same classloader that loaded the class to load something else for you. Like so:

InputStream is = getClass().getClassLoader().getResourceAsStream("com/foo/something.xml");

[himanshu.harish agrawal]

Below is a working solution for loading a file/XSD from JAR at runtime. Also, to cross check if it worked you can print the content on console.

                     try {
				InputStream is = null;
				BufferedReader br = null;
				String line;
				ArrayList list = new ArrayList();

				try { 
				  is = RequestXMLValidator.class.getResourceAsStream("/schemas/wac-querySubscriber.xsd");
				  br = new BufferedReader(new InputStreamReader(is));
				  while (null != (line = br.readLine())) {
				     list.add(line);
				  }
				}
				catch (Exception e) {
				  e.printStackTrace();
				}
				
				Iterator it = list.iterator();
                                System.out.println(" ################Printing Starts ##################### ");
				while(it.hasNext()) {
				     System.out.println(it.next());
				}
			} catch (Exception e) {
				
				e.printStackTrace();
			}

In the above code "RequestXMLValidator" is the name of the class, so just replace it with you class name and "/schemas/wac-querySubscriber.xsd" is the path of the file/XSD in JAR.

In case, if you parse the XSD and prepare your Document object then use below line of code instead of creating a BufferedReader object and so on.

Document document = dBuilder.parse(inputStream);

[Campbell Ritchie]

Welcome to the Ranch

Why have you posted two apparently identical answers on rather old posts?

Please use the code button; since you are new I shall edit your post (maybe both) and you san see ho wmuch easier it is to read.

[himanshu.harish agrawal]

Thanks Campbell !!!

[himanshu.harish agrawal]