Uploading a file to server using servlet (Without using Jakarta Commons)

Hi,
I was trying to upload a file to server using servlet, but i need to do that without the help of anyother API packages like Jakarta Commons Upload. If any class for retrieval is necessary, how can i write my own code to upload from client machine?.
From
Velu

<p>Why put such a restriction on the solution? Whats wrong about using that library?
The uploading bit is easy - you put a <input type="file"> component on the form, and set it to be method="post" and enctype="multipart/form-data"
Reading the input stream at the other end - thats harder - which is why they wrote a library for it. </p>
why i gave the restriction is that, i have a question that <code>'can't we implement the same upload'</code>
I was with the view that the same can be implemented by our own code right?

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    But when iam trying to access it through other system it is not doing so
    Giving an internal server error i,e 500.
    Iam putting the necessary documents for ur reference.
    Plz help me soon .
    My exact requirement is i have to upload a file to the upload folder located in the server.
    And i have to get the path of that file and display the file path exactly below the browse button from where iam uploaded a file.
    That should be done without page refresh and submit thats y iam used Ajax
    Any help would greatly appreciated
    Thanks and Regards
    Meerasaaheb.
    The action class is FilePathAction
    package actions;
    import org.apache.struts.action.Action;
    import org.apache.struts.action.ActionForm;
    import org.apache.struts.action.ActionForward;
    import org.apache.struts.action.ActionMapping;
    import javax.servlet.ServletException;
    import javax.servlet.http.HttpServletRequest;
    import javax.servlet.http.HttpServletResponse;
    import java.io.*;
    public class FilePathAction extends Action{
    public ActionForward execute(ActionMapping mapping, ActionForm form,
    HttpServletRequest request, HttpServletResponse response)
    throws IOException, ServletException
    String contextPath1 = "";
    String uploadDirName="";
    String filepath="";
    System.out.println(contextPath1 );
    String inputfile = request.getParameter("filepath");
    uploadDirName = getServlet().getServletContext().getRealPath("/upload");
    File f=new File(inputfile);
    FileInputStream fis=null;
    FileOutputStream fo=null;
    File f1=new File(uploadDirName+"/"+f.getName());
    fis=new FileInputStream(f);
    fo=new FileOutputStream(f1);
    try
    byte buf[] = new byte[1024*8]; /* declare a 8kB buffer */
    int len = -1;
    while((len = fis.read(buf)) != -1)
    fo.write(buf, 0, len);
    catch(Exception e)
    e.printStackTrace();
    filepath=f1.getAbsolutePath();
    request.setAttribute("filepath", filepath);
    return mapping.findForward("filepath");
    the input jsp is
    filename.jsp
    <%@ page language="java" contentType="text/html; charset=ISO-8859-1"
    pageEncoding="ISO-8859-1"%>
    <!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
    <html>
    <head>
    <meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
    <title>Insert title here</title>
    <script type="text/javascript">
    alertflag = false;
    var xmlHttp;
    function startRequest()
    if(alertflag)
    alert("meera");
    xmlHttp=createXmlHttpRequest();
    var inputfile=document.getElementById("filepath").value;
    xmlHttp.open("POST","FilePathAction.do",true);
    xmlHttp.onreadystatechange=handleStateChange;
    xmlHttp.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');
    xmlHttp.send("filepath="+inputfile);
    function createXmlHttpRequest()
    //For IE
    if(window.ActiveXObject)
    xmlHttp=new ActiveXObject("Microsoft.XMLHTTP");
    //otherthan IE
    else if(window.XMLHttpRequest)
    xmlHttp=new XMLHttpRequest();
    return xmlHttp;
    //Next is the function that sets up the communication with the server.
    //This function also registers the callback handler, which is handleStateChange. Next is the code for the handler.
    function handleStateChange()
    var message=" ";
    if(xmlHttp.readyState==4)
    if(alertflag)
    alert(xmlHttp.status);
    if(xmlHttp.status==200)
    if(alertflag)
    alert("here");
    document.getElementById("div1").style.visibility = "visible";
    var results=xmlHttp.responseText;
    document.getElementById('div1').innerHTML = results;
    else
    alert("Error loading page"+xmlHttp.status+":"+xmlHttp.statusText);
    </script></head><body><form name="thumbs" enctype="multipart/form-data" method="post" action="">
    <input type="file" name="filepath" id="filepath" onchange="startRequest();"/>
    </form>
    <div id="div1" style="visibility:hidden;">
    </div></body></html>
    The ajax response is catching in a dummy.jsp
    <%=(String)request.getAttribute("filepath")%>
    corresponding action mapping
    <action path="/FilePathAction" type="actions.FilePathAction">
    <forward name="filepath" path="/dummy.jsp"/>
    </action>
    So plz help me to upload a file to the server from any PC.
    Iam searched alot but didnt get any solution.

    Plz help me soon if it possible so
    Iam in great need.
    I have worked alot but not worked out.
    Any help greatly appreciated

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