Multi part upload file servlet [duplicate]

Answer 1

This does not work for IE7 and below. Apparently you need to add another attribute to your form encoding ="multipart/form-data"

Answer 2

To deal with enctype="multipart/form-data" we can not use request.getParameter() directly

Now to deal with the problem

Now, for uploading a file to the server, there can be various ways. But, I am going to use MultipartRequest class provided by oreilly. For using this class you must have cos.jar file.

public class UploadServlet extends HttpServlet 
{  

    public void doPost(HttpServletRequest request, HttpServletResponse response)                     throws ServletException, IOException 
    {           
        MultipartRequest m=new MultipartRequest(request,"d:/new");  
        out.print("successfully uploaded");  
    }  
}  

this will upload your file to d:/new

Now to retrive parameter of multipart request you have to use FilenameUtils class and getOriginalFileName() method of MultipartRequest class.

String file = FilenameUtils.getName(req.getOriginalFileName("myfile"))+"\\";
String message = req.getParameter("message");

Answer 3

Step 1

Read adatapost's post.

Step 2

Check out the Apache Commons FileUpload project.

There's a similarly workable solution by O'Reily, but its license of use requires you buy a book, and even that requirement is so poorly articulated that I won't benefit it with yet another link.

Answer 4

To browse and select a file for upload you need a <input type="file"> field in the form. As stated in the HTML specification you need to use the POST method and the enctype attribute of the form has to be set to multipart/form-data.

<form action="uploadServlet" method="post" enctype="multipart/form-data">
    <input type="file" name="file" />
    <input type="submit" />
</form>

After submitting such a form the form data is available in multipart format in the HttpServletRequest#getInputStream(). For testing(!) purposes you can read the stream using the following snippet:

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

You however need to parse the stream byte by byte (instead of char by char). Prior to the fresh new Servlet 3.0 API, the standard Servlet API didn't provide any builtin facilities to parse them. The normal form fields are also not available the usual request.getParameter() way, they are included in the multipart form data stream.

If you're not on Servlet 3.0 yet (which is only bit less than 2 monts old), then you need to parse the stream yourself. Parsing such a stream requires precise background knowledge of how multipart form data requests are specified and structured. To create a perfect multipart parser you'll have to write a lot of code. But fortunately there's the Apache Commons FileUpload which has proven its robustness with years. Carefully read both the User Guide and Frequently Asked Questions to find code examples and learn how to use it to an optimum degree (take MSIE into account!).

Answer 5

Step-1

set enctype form tag attribute.

<form enctype="multipart/form-data" ....>
   <input type="file" id="file1" name="file"/>
   .... other stuff
</form>

Step-2

Read Justin's post.