I have a jar library which contains images, css, etc.. This jar is based on a maven project. All files are placed in src/main/resources/web directory.
The jar is placed in shared/lib of Tomcat 5.5 in order to be shared to all web applications.
Now, in a standard web application, how to access to a css (for example) into stylesheet link of a JSP page ?
Thank you.
nota: with Zk framework, we can access with a simple linke i.e "~./web/css/header.css" but it is not working with JSP :(
You can but you need some manual coding, as default servlet (part of container /tomcat/ that serves static resources) does know nothing about your jars. You need to
I implemented simple prototype, it does handle only CSS files and it does not cover any corner cases. But it works and you can extend it as you need:
Servlet is simple, it just takes part or URL behind its mapping and loads resource from classpath (e.g. jar):
package cz.literak.demo; import javax.servlet.ServletException; import javax.servlet.http.HttpServlet; import javax.servlet.http.HttpServletRequest; import javax.servlet.http.HttpServletResponse; import java.io.IOException; import java.io.InputStream; import java.io.PrintWriter; public class JarServlet extends HttpServlet { @Override protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { String path = request.getPathInfo(); setContentType(path, response); InputStream streamIn = null; try { streamIn = getClass().getResourceAsStream(path); PrintWriter writer = response.getWriter(); int c; while ((c = streamIn.read()) != -1) { writer.write(c); } } catch (IOException e) { streamIn.close(); } } private void setContentType(String path, HttpServletResponse response) { if (path.toLowerCase().endsWith(".css")) { response.setContentType("text/css"); } // TODO other mime types } } You need to register the servlet and its mapping:
<servlet> <servlet-name>JarDefault</servlet-name> <servlet-class>cz.literak.demo.JarServlet</servlet-class> </servlet> <servlet-mapping> <servlet-name>JarDefault</servlet-name> <url-pattern>/jar/*</url-pattern> </servlet-mapping> And you can use it this way:
<link rel="stylesheet" type="text/css" href="jar/styles/layout.css"/> I copied file layout.css under directory styles in one jar that is part of my war. Easy, is not it?
The problem was that I was going a step too far in calling the parse method of XMLReader. The parse method accepts an InputSource, so there was no reason to even use a FileReader. Changing the last line of the code above to
xr.parse( new InputSource( filename )); Another one sample code: Sometimes it is useful to distribute an application in a jar file through Java Web Start or any other way. So, you could have to read some resource (images or properties file) from inside a jar. How can you do it ? It’s very simple, here’s an example to retrieve an image:
ImageIcon image = (new ImageIcon(getClass().getResource("yourpackage/mypackage/image.gif"))); In general, you can retrieve an InputStream in the following way:
InputStream is = this.getClass().getClassLoader().getResourceAsStream("yourpackage/mypackage/myfile.xml"); 