Convenient way to parse incoming multipart/form-da

问题:

This question already has an answer here:

• How to upload files to server using JSP/Servlet? 12 answers

Is there any convenient way to read and parse data from incoming request.

E.g client initiate post request

URLConnection connection = new URL(url).openConnection();
connection.setDoOutput(true);
connection.setRequestProperty(\"Content-Type\", \"multipart/form-data; boundary=\" + boundary);
PrintWriter writer = null;
try {
    OutputStream output = connection.getOutputStream();
    writer = new PrintWriter(new OutputStreamWriter(output, charset), true); // true = autoFlush, important!
    // Send normal param.
    writer.println(\"--\" + boundary);
    writer.println(\"Content-Disposition: form-data; name=\\\"param\\\"\");
    writer.println(\"Content-Type: text/plain; charset=\" + charset);
    writer.println();
    writer.println(param);

I’m not able to get param using request.getParameter(\"paramName\"). The following code

BufferedReader reader = new BufferedReader(new InputStreamReader(
    request.getInputStream()));
  StringBuilder sb = new StringBuilder();
  for (String line; (line = reader.readLine()) != null;) {
   System.out.println(line);

  }

however displays the content for me

-----------------------------29772313742745
Content-Disposition: form-data; name=\"name\"
J.Doe
-----------------------------29772313742745
Content-Disposition: form-data; name=\"email\"
[e-mail]
-----------------------------29772313742745

What is the best way to parse incoming request? I don’t want to write my own parser, probably there is a ready solution.

回答1:

multipart/form-data encoded requests are indeed not by default supported by the Servlet API prior to version 3.0. The Servlet API parses the parameters by default using application/x-www-form-urlencoded encoding. When using a different encoding, the request.getParameter() calls will all return null. When you\'re already on Servlet 3.0 (Glassfish 3, Tomcat 7, etc), then you can use HttpServletRequest#getParts() instead. Also see this blog for extended examples.

Prior to Servlet 3.0, a de facto standard to parse multipart/form-data requests would be using Apache Commons FileUpload. Just carefully read its User Guide and Frequently Asked Questions sections to learn how to use it. I\'ve posted an answer with a code example before here (it also contains an example targeting Servlet 3.0).

回答2:

Solutions:

Solution A:

1. Download http://www.servlets.com/cos/index.html
2. Invoke getParameters() on com.oreilly.servlet.MultipartRequest

Solution B:

1. Download http://jakarta.Apache.org/commons/fileupload/
2. Invoke readHeaders() in org.apache.commons.fileupload.MultipartStream

Solution C:

1. Download http://users.boone.net/wbrameld/multipartformdata/
2. Invoke getParameter on com.bigfoot.bugar.servlet.http.MultipartFormData

Solution D:

Use Struts. Struts 1.1 handles this automatically.

Reference: http://www.jguru.com/faq/view.jsp?EID=1045507

回答3:

Not always there\'s a servlet before of an upload (I could use a filter for example). Or could be that the same controller ( again a filter or also a servelt ) can serve many actions, so I think that rely on that servlet configuration to use the getPart method (only for Servlet API >= 3.0), I don\'t know, I don\'t like.

In general, I prefer independent solutions, able to live alone, and in this case http://commons.apache.org/proper/commons-fileupload/ is one of that.

List<FileItem> multiparts = new ServletFileUpload(new DiskFileItemFactory()).parseRequest(request);
    for (FileItem item : multiparts) {
        if (!item.isFormField()) {
            //your operations on file
        } else {
            String name = item.getFieldName();
            String value = item.getString();
            //you operations on paramters
        }
}