Ajax call to servlet to send form

Question

Am unable to send form to my servlet, am not sure what am doing wrong, my servlet name is SupportReq,

var formdata = new FormData();
           if($scope.files.length>0){
           formdata.append("attachment0",$scope.files[0]);
           }
           formdata.append("description",$scope.rsp.observations);
           formdata.append("title",$scope.rsp.title);
           $scope.rsp.severity = null;
           $scope.rsp.urgency = 'N';
           formdata.append("secerity",$scope.rsp.severity);
           formdata.append("urgency",$scope.rsp.urgency);
           formdata.append("reviewDate",$scope.date);
           formdata.append("subtopic",subtopictest.id);            

           $.ajax({
                url: 'SupportRequest',
                data: $(".formdata").serialize(),
                contentType: 'multipart/form-data',
                processData: false,
                type: 'POST',
                success: function(data){
                    console.log(data);
                }
            });

What might be wrong here?

This is my web.xml

<servlet>
<servlet-name>SupportRequest</servlet-name>
<display-name>SupportRequest</display-name>
<description></description>
<servlet-class>com.request.SupportRequestCreatorServlet</servlet-class>

Ok ... take a look at : https://developer.mozilla.org/en-US/docs/Web/API/FormData/Using_FormData_Objects — Abdennour TOUMI, Jul 21 '16 at 05:53

score 0 · Answer 1 · edited May 23 '17 at 11:58

0

This line in ajax call :

data: $(".formdata").serialize(),

Should be :

   data:formdata,

For more info , check here .

__

Useful DEMO

edited May 23 '17 at 11:58

Community

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answered Jul 21 '16 at 05:55

Abdennour TOUMI

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But it doesnt seem to work cause i get an error: org.apache.commons.fileupload.FileUploadException: the request was rejected because no multipart boundary was found – Kiran Kumar Jul 21 '16 at 06:26

Ajax call to servlet to send form

1 Answers1