Ajax does not pass information to servlet

1

I have a problem with the code, the servlet is not called. Only works if I put action="FileUploader" in form. Is there a mistake? Do you have any other way to do it? What I need: An array of images, where images are received in a single multiple / form data field (that is, all images received at one time) that is passed to the servlet, without redirecting the page. Thank you in advance.

<form id="form1">    

    <label for="sampleText">Please enter a text</label>

    <input id="sampleText" name="sampleText" type="text" /> <br/>

    <label for="sampleFile">Please select a file

    <input id="sampleFile" name="sampleFile" type="file" /> <br/>

    <input id="uploadBtn" type="button" value="Ajax Submit" onClick="performAjaxSubmit();"></input>

</form>

<script type="text/javascript">

    function performAjaxSubmit() {

        var sampleText = document.getElementById("sampleText").value;

        var sampleFile = document.getElementById("sampleFile").files[0];

        var formdata = new FormData();

        formdata.append("sampleText", sampleText);

        formdata.append("sampleFile", sampleFile);

        var xhr = new XMLHttpRequest();       

        xhr.open("POST","FileUploader", true);

        xhr.send(formdata);

        xhr.onload = function(e) {

            if (this.status == 200) {

               alert(this.responseText);

            }

        };                    

    }   

</script>


public class FileUploader extends HttpServlet {

    protected void doPost(HttpServletRequest request,

                          HttpServletResponse response)

                          throws ServletException, IOException {

        String ajaxUpdateResult = "";

        try {

            List items = new ServletFileUpload(new DiskFileItemFactory()).parseRequest(request);            

            for (FileItem item : items) {

                if (item.isFormField()) {

                    ajaxUpdateResult += "Field " + item.getFieldName() + 

                    " with value: " + item.getString() + " is successfully read\n\r";

                } else {

                    String fileName = item.getName();

                    InputStream content = item.getInputStream();

                    response.setContentType("text/plain");

                    response.setCharacterEncoding("UTF-8");

                    // Do whatever with the content InputStream.

                    System.out.println(Streams.asString(content));

                    ajaxUpdateResult += "File " + fileName + " is successfully uploaded\n\r";

                }

            }

        } catch (FileUploadException e) {

            throw new ServletException("Parsing file upload failed.", e);

        }

        response.getWriter().print(ajaxUpdateResult);

    }
}
    
asked by anonymous 09.04.2016 / 22:53

0 answers