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Manish Kumar
File Uploading Problem
0 Answer(s)      2 years and 2 months ago
Posted in : Java Beginners


I have a file uploading code but it create problem

$(document).ready(function(){
$('#upload').click(function(){

var fileName=$('#myFile').val();
alert(fileName);
$.ajax({
  type: 'POST',
  dataType: 'json',
  url: 'myServlet',
  data:{path:fileName},
  contentType:'multipart/form-data'
  success: function(msg){
    //alert( "Data Saved: " + msg );
    document.getElementById('filenameSeq').innerHTML='<div style="width:200px; color: black;background-color: white;"><ul><li>'+msg+'&nbsp&nbsp;<a href="#####"><b>X</b></a></li></ul></div>';
  }
});
});

});

but when we run it the code it gives the problem

org.apache.commons.fileupload.FileUploadException: the request was rejected because no multipart boundary was found

and my server side code is

protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        try{ 
            System.out.println("------------------------------------------");
            FileUpload fup=new FileUpload(); 
            boolean isMultipart = FileUpload.isMultipartContent(request); 
            // Create a new file upload handler 
            System.out.println(isMultipart); 
            DiskFileUpload upload = new DiskFileUpload(); 

            // Parse the request 
            List /* FileItem */ items = (List) upload.parseRequest(request); 

            Iterator iter = (Iterator) ((java.util.List) items).iterator(); 
            while (((java.util.Iterator) iter).hasNext()) { 

            FileItem item = (FileItem) ((java.util.Iterator) iter).next(); 

            if (item.isFormField()) { 
            System.out.println("its a field"); 
            } else { 
            System.out.println("its a file"); 
            System.out.println(item.getName()); 
            File cfile=new File(item.getName()); 
            File tosave=new File(getServletContext().getRealPath("/Web-INF/manu/"),cfile.getName()); 
            System.out.println(tosave);
            item.write(tosave); 
            } 
            } 

            }catch(Exception e){System.out.println(e);} 
    }               

this code run perfectly well when i send action without jquery

if u have any solution then share it


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