upload file from a client machine to server machine

[praveenkumarvpk]

Hi friends Please help me!

Following is my servlet-code

import java.util.Enume ration;
import javax.servlet.* ;
import javax.servlet.h ttp.*;
import java.io.*;

import com.oreilly.ser vlet.MultipartR equest;
import com.oreilly.ser vlet.MultipartR esponse;
import com.oreilly.ser vlet.multipart. DefaultFileRena mePolicy;

public class DemoUpload extends HttpServlet {
private String dirName;

public void init(ServletCon fig config) throws ServletExceptio n {
super.init(conf ig);
// read the uploadDir from the servlet parameters
dirName = config.getInitP arameter("uploa dDir");
dirName="c:/a";//SPORTS.JPEG";

//System.out.prin tln("before :"+dirName);

if (dirName == null) {
throw new ServletExceptio n("Please supply uploadDir parameter");
}
}

public void doGet(HttpServl etRequest request, HttpServletResp onse response)
throws ServletExceptio n, IOException {
PrintWriter out = response.getWri ter();
response.setCon tentType("text/plain");
out.println("De mo Upload Servlet using MultipartReques t");
out.println();

try {
// Use an advanced form of the constructor that specifies a character
// encoding of the request (not of the file contents) and a file
// rename policy.
System.out.prin tln("before :"+dirName);
MultipartReques t multi =new MultipartReques t(request, dirName, (10*1024*1024), "ISO-8859-1", new DefaultFileRena mePolicy());

System.out.prin tln("After :"+dirName);
out.println("PA RAMS:");
Enumeration params = multi.getParame terNames();
while (params.hasMore Elements()) {
String name = (String)params. nextElement();
String value = multi.getParame ter(name);
out.println(nam e + "=" + value);
}
out.println();

out.println("FI LES:");
Enumeration files = multi.getFileNa mes();
while (files.hasMoreE lements()) {
String name = (String)files.n extElement();
String filename = multi.getFilesy stemName(name);
String originalFilenam e = multi.getOrigin alFileName(name );
String type = multi.getConten tType(name);
File f = multi.getFile(n ame);
out.println("na me: " + name);
out.println("fi lename: " + filename);
out.println("or iginalFilename: " + originalFilenam e);
out.println("ty pe: " + type);
if (f != null) {
out.println("f. toString(): " + f.toString());
out.println("f. getName(): " + f.getName());
out.println("f. exists(): " + f.exists());
out.println("f. length(): " + f.length());
}
out.println();
}
}
catch (IOException lEx) {
this.getServlet Context().log(l Ex, "error reading or saving file");

System.out.prin tln("HERE THE EXCEPTION IS RAISING : "+lEx);
}
}
}


When i execute this i'm getting an exception that


before :c:/a
Feb 2, 2008 12:17:50 PM org.apache.cata lina.core.Appli cationContext log
SEVERE: error reading or saving file
java.io.IOExcep tion: Posted content type isn't multipart/form-data
at com.oreilly.ser vlet.multipart. MultipartParser .<init>(Multipa rtParser.java:1 66)
at com.oreilly.ser vlet.MultipartR equest.<init>(M ultipartRequest .java:222)
at DemoUpload.doGe t(DemoUpload.ja va:38)


Please help to upload a file

[praveenkumarvpk]

Hey guys Please help to find it.... urgent

Thank U

[r035198x]

Read your exception trace. It's telling you exactly what the problem is.