Access file in WebContent folder from a servlet

Question

I'm trying to generate a PDF document using FOP. The pdf generation code is kept in a servlet and the xsl is in a specific folder in the WebContent folder.

How can I access this xsl file by giving a relative path? It works only if I give the complete path in the File object.

I need to generate the xml content dynamically. How can I give this dynamically generated xml as the source instead of a File object?

Please provide your suggestions.

Answer 1

I used the following method to read the file under web content

BufferedReader reader = new BufferedReader(new InputStreamReader(request.getSession().getServletContext().getResourceAsStream("/json/sampleJson.json")));

Now all the file content is available in the reader object.

Answer 2

For a direct and independent container implementation, you can access the resourcewith the following method getResource() inside your servlet:

/start servlet/

public InputStream getResource(String resourcePath) {
  ServletContext servletContext = getServletContext();
  InputStream openStream = servletContext.getResourceAsStream( resourcePath );
  return openStream;
}

public void testConsume() {
  String path = "WEB-INF/teste.log";
  InputStream openStream = getResource( path );

  int c = -1;
  byte[] bb = new byte[1024];
  while ( -1 != ( c = openStream.read( bb ) ) ) {
    /* consume stream */
  }
  openStream.close();
}

/end servlet/

Answer 3

To get the path you can just do:

String path = s.getServletContext().getRealPath("/WEB-INF/somedir/hdfeeh");         

s is the class that implements HTTPServlet.You can also use this.getServletContext() if its your servlet class.

Then pass this as a parameter.

As far as using dynamically generated XML, the library you're using should support using an input stream, write your XML, convert it to a byte array, then wrap it in a ByteArrayInputStream and use this.