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POUsing Apache commons FileUpload
 

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  1. POUsing Apache commons FileUpload
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    <p>This just won't work. The problem is that I do not know enough to even know what is supposed to happen. I can't debug this code. I'd like to store upload to temporary folder "temp" and then to move them to "applets". Please help? The servlet is obviously being accessed, but I can't find the uploaded files... Thanks in advance. </p> <p>Form (which is created using a scriptlet - I put this here if that could cause problems): </p> <pre><code>&lt;% out.write("&lt;p&gt;Upload a new game:&lt;/p&gt;"); out.write("&lt;form name=\"uploadForm\" action=\"game.jsp\" " + "method=\"POST\" enctype=\"multipart/form-data\"&gt;" + "&lt;input type=\"file\" name=\"uploadSelect\" value=\"\" width=\"20\" /&gt;" + "&lt;br&gt;&lt;input type=\"submit\" value=\"Submit\" name=\"uploadSubmitButton\" " + "onclick = \"submitToServlet2('UploadGameServlet');\"&gt;" + "&lt;/form&gt;"); %&gt; </code></pre> <p>Which calls this javascript:</p> <pre><code>function submitToServlet2(newAction) { document.uploadForm.action = newAction; } </code></pre> <p>Which in turn goes to the servlet (code included in full, since there may be some important element hiding)</p> <pre><code>package org.project; import java.io.*; import java.util.Iterator; import java.util.List; import java.util.logging.Level; import java.util.logging.Logger; // import servlet stuff import org.apache.commons.fileupload.*; public class UploadGameServlet extends HttpServlet { /** * Processes requests for both HTTP &lt;code&gt;GET&lt;/code&gt; and &lt;code&gt;POST&lt;/code&gt; methods. * @param request servlet request * @param response servlet response */ protected void processRequest(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { response.setContentType("text/html;charset=UTF-8"); if (ServletFileUpload.isMultipartContent(request)) { try { // Create a factory for disk-based file items FileItemFactory factory = new DiskFileItemFactory(); // Create a new file upload handler ServletFileUpload upload = new ServletFileUpload(factory); // Parse the request List items = upload.parseRequest(request); /* FileItem */ File repositoryPath = new File("\\temp"); DiskFileItemFactory diskFileItemFactory = new DiskFileItemFactory(); diskFileItemFactory.setRepository(repositoryPath); Iterator iter = items.iterator(); while (iter.hasNext()) { FileItem item = (FileItem) iter.next(); File uploadedFile = new File("\\applets"); item.write(uploadedFile); } } catch (FileUploadException ex) { Logger.getLogger(UploadGameServlet.class.getName()).log(Level.SEVERE, null, ex); } catch (Exception ex) { Logger.getLogger(UploadGameServlet.class.getName()).log(Level.SEVERE, null, ex); } } PrintWriter out = response.getWriter(); try { out.println("&lt;html&gt;"); out.println("&lt;head&gt;"); out.println("&lt;title&gt;Servlet UploadGameServlet&lt;/title&gt;"); out.println("&lt;/head&gt;"); out.println("&lt;body&gt;"); out.println("&lt;h1&gt;Servlet UploadGameServlet at " + request.getContextPath () + "&lt;/h1&gt;"); out.println("&lt;/body&gt;"); out.println("&lt;/html&gt;"); } finally { out.close(); } } </code></pre> <p>}</p>
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    1. COCan't you debug the servlet code? Are you sure the servlet is being accessed? I don't know if you need to return **true** in the submitToServlet2 function in order to submit the request.
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      1. USMiguel Ping
    2. COYes, the server outputs a heading... So I know it at least gets to the output lines. How can I debug the servlet code?
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      1. USpypmannetjies
    3. COYou can try to add some logging statements inside your if. Also, are you getting any exceptions logged?
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      1. USSteve K
 

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