java - how to get new request object after ajax called -



java - how to get new request object after ajax called -

i developing web project using jsp , servlet.

i trying upload file using ajax called. file uploaded. but, while ajax called controller (servlet file) send request , response object jsp file.

jsp file

$(document).ready(function(){ $(':file').change(function(){ var fileobj = this.files[0]; var form = $('#mobj'); var fd = new formdata(); fd.append( 'file', fileobj); $.ajax({ url:form.attr('action'), type:form.attr('method'), data:fd, processdata: false, contenttype: false, async:false, }).done(function(){ alert('ajax complete');

//////////////////////////////////////////////////////////////// here getting old request value. want new request , response object after ajax called.

var check2bool = <%=context.getattribute("comefromuploadtemp")%>; var check2 = <%=request.getattribute("comefromuploadtemp")%>; alert(check2bool + " " + check2); }).fail(function() { alert( "error" ); $('#ldiv').hide(); }); });

servlet

protected void doget(httpservletrequest request, httpservletresponse response) throws servletexception, ioexception { servletcontext context = getservletcontext(); /** * display xml file user */ templatevo template = null; templateutil util = new templateutil(); //prepare xml file path string path=(string) context.getattribute(constants.user_dir); string strfilepath = null; //upload xml file if(servletfileupload.ismultipartcontent(request)){ try{ list<fileitem> multiparts = new servletfileupload(new diskfileitemfactory()).parserequest(request); for(fileitem item : multiparts){ if(!item.isformfield()){ string name = new file(item.getname()).getname(); request.setattribute("filename",name); string path1 = path + file.separator + "data-in"; strfilepath = fileutil.createfileonfilesystem(item,path1,name,"xml"); } } }catch(exception ex){ ex.printstacktrace(); } } //set variables request object template = util.readxml(strfilepath); context.setattribute("comefromuploadtemp", true); request.setattribute("comefromuploadtemp", true); request.getrequestdispatcher("mappingobject.jsp").forward(request, response); } protected void dopost(httpservletrequest request, httpservletresponse response) throws servletexception, ioexception { doget(request, response); }

the request jsp workspace, notajax. u should pass info through response i.o. request. like:

protected void doget(httpservletrequest request, httpservletresponse response) throws servletexception, ioexception { //... code here response.getoutputstream().write(data); }

then u can parse response , info want.

http://www.coderanch.com/t/522069/servlets/java/adding-custom-data-response-headers http://www.w3schools.com/ajax/ajax_xmlhttprequest_response.asp

java jquery ajax jsp servlets

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