How can my Servlet receive parameters from a multipart/form-data form?

Question

I have a page that has this piece of code:

<form action="Servlet" enctype="multipart/form-data">
<input type="file" name="file">
<input type="text" name="text1">
<input type="text" name="text2">
</form>

When I use request.getParameter("text1"); in my Servlet it shows null. How can I make my Servlet receive the parameters?

Possible duplicate of [How to upload files in JSP/Servlet?](http://stackoverflow.com/questions/2422468/how-to-upload-files-in-jsp-servlet/2424824#2424824), in other words: just by the same API as you retrieved the file. — BalusC, Mar 12 '12 at 13:13

score 6 · Answer 1 · answered Mar 12 '12 at 12:45

6

All the request parameters are embedded into the multipart data. You'll have to extract them using something like Commons File Upload: http://commons.apache.org/fileupload/

answered Mar 12 '12 at 12:45

gorjusborg

656
4
18

score 2 · Answer 2 · edited Jul 15 '13 at 14:01

2

Pleepleus is right, commons-fileupload is a good choice.
If you are working in servlet 3.0+ environment, you can also use its multipart support to easily finish the multipart-data parsing job. Simply add an @MultipartConfig on the servlet class, then you can receive the text data by calling request.getParameter(), very easy.

Tutorial - Uploading Files with Java Servlet Technology

edited Jul 15 '13 at 14:01

Aboutblank

697
3
14
31

answered Jul 15 '13 at 13:35

qiangbro

128
1
9

score 2 · Answer 3 · answered Mar 12 '12 at 12:45

2

Use getParts()

answered Mar 12 '12 at 12:45

Francisco Paulo

6,284
26
25

From what class is getParts() from? – Bolaum Mar 12 '12 at 12:49
Ooooh, underrated answer! – Nathan Crause Feb 22 '22 at 19:21

score 1 · Answer 4 · answered Nov 07 '13 at 15:31

You need to send the parameter like this:

writer.append("--" + boundary).append(CRLF);
writer.append("Content-Disposition: form-data; name=\"" + urlParameterName + "\"" )
                .append(CRLF);
writer.append("Content-Type: text/plain; charset=" + charset).append(CRLF);
writer.append(CRLF);
writer.append(urlParameterValue).append(CRLF);
writer.flush();

And on servlet side, process the Form elements:

items = upload.parseRequest(request);
Iterator iter = items.iterator();
while (iter.hasNext()) {
       item = (FileItem) iter.next();
       if (item.isFormField()) {
          name = item.getFieldName(); 
          value = item.getString();

   }}

How can my Servlet receive parameters from a multipart/form-data form?

4 Answers4

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