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From David Smith <d...@cornell.edu>
Subject Re: where to store user-generated files?
Date Wed, 21 Feb 2007 15:41:48 GMT
I'm sure this is horribly insufficient, so I would recommend reading the
servlet spec.  It's not all that bad a read as far as specifications go
and you can learn a lot about how tomcat operates.

If you have a servlet mapping all *.jpg to some servlet "myServlet" in
web.xml, web.xml also has another section where the servlets are
individually declared.  In there you'll find the class that backs the
servlet named "myServlet".  All that means is the request for the
resource goes to that specific servlet class to be handled.  What the
servlet does with it is open-ended.  In the case of the default servlet,
it basically calls sc.getResourceAsStream( request.getServletPath() +
request.getPathInfo() ) and returns the contents of the file to the
client.  That's over simplified -- the real default servlet does some
security checks to be sure it's not returning something it shouldn't and
set's the mime type in the response, but you get the idea. 
sc.getResourceAsStream is aware of the webapp's location in the
filesystem and takes care of finding the resource in the webapp's folder.

sc.getRealPath will only return a full path to a file if the webapp is
NOT compressed in a web archive file (.war).  From within a .war file,
it always returns null and isn't recommended except in rare cases where
you might want to write access.  This is all in the spec.

Servlet 2.4 spec is available at
http://jcp.org/aboutJava/communityprocess/final/jsr154/index.html

--David

Williams, Allen wrote:

>I'm new at this, so bear with me here for a moment...
>
>The servlet mapping seems to me to tell tomcat "anytime you have a
>request for a URI with .jpg extension, deliver the request to this
>servlet", but that doesn't give any information about where in the
>"real" file system said jpeg is stored, does it?  So, when you call
>sc.getRealPath(), how does the servlet context know where to go?
>Doesn't there have to be a mapping or alias somewhere (server.xml,
>web.xml,...?) that resolves, or translates "ThisTypeofFileName.ext" into
>"/real/path/in/OS/ThatTypeOfFileName.oxt"?
>
>
>-----Original Message-----
>From: John Pedersen [mailto:john.pedersen@gmail.com] 
>Sent: Wednesday, February 21, 2007 4:37 AM
>To: Tomcat Users List
>Subject: Re: where to store user-generated files?
>
>Looks like roll your own then!
>
>A few thoughts on the matter - maybe someone could add to them?
>
>It should be easy to map requests for images to a servlet, which can
>then find the appropriate image file wherever it might be ( within or
>outside the server ). Like this in the web.xml file:
>
><servlet-mapping>
>	<servlet-name>servletName</servlet-name>
>	<url-pattern>*.jpg</url-pattern>
></servlet-mapping>
>
>?
>
>But how is the image then added to the reponse? Another servlet (
>behind the scenes - I am actually using  the Spring framework ) is
>handling the request/response. Can the request/response be passed to
>the image-providing servlet for the images within a page to be written
>to the reponse in this kind of way:
>
> // This method is called by the servlet container to process a GET
>request.
>    public void doGet(HttpServletRequest req, HttpServletResponse
>resp) throws IOException {
>        // Get the absolute path of the image
>        ServletContext sc = getServletContext();
>        String filename = sc.getRealPath("image.gif");
>
>        // Get the MIME type of the image
>        String mimeType = sc.getMimeType(filename);
>        if (mimeType == null) {
>            sc.log("Could not get MIME type of "+filename);
> 
>resp.setStatus(HttpServletResponse.SC_INTERNAL_SERVER_ERROR);
>            return;
>        }
>
>        // Set content type
>        resp.setContentType(mimeType);
>
>        // Set content size
>        File file = new File(filename);
>        resp.setContentLength((int)file.length());
>
>        // Open the file and output streams
>        FileInputStream in = new FileInputStream(file);
>        OutputStream out = resp.getOutputStream();
>
>        // Copy the contents of the file to the output stream
>        byte[] buf = new byte[1024];
>        int count = 0;
>        while ((count = in.read(buf)) >= 0) {
>            out.write(buf, 0, count);
>        }
>        in.close();
>        out.close();
>    }
>
>( from http://www.exampledepot.com/egs/javax.servlet/GetImage.html )
>
>I'm off for a walk to mull it over - any suggestions while I am out
>and before I get to experimenting will be most welcome.
>
>Thanks,
>
>John
>
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>  
>


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