RESTlet: How to process multipart/form-data requests?

问题

How do you catch incoming @Post variables when it is a multipart/form-data request?

For a regular Post request I would do:

@Post
public void postExample(Representation entity) throws Exception{
   Form form = new Form(entity);
   System.out.println(form.getFirstValue("something"));
}

But because it is a multipart/form-data request the above outputs null

I'm a Java newbie so be gentle :)

PS: I'm not interested in processing the incoming files, just the text fields.

回答1:

This is a paste from one of my methods (Restlet 2.0). Here I have a form that includes one file upload plus other fields, therefore it is rather complete:

@Post
public Representation createTransaction(Representation entity) {
    Representation rep = null;
    if (entity != null) {
        if (MediaType.MULTIPART_FORM_DATA.equals(entity.getMediaType(), true)) {
            // 1/ Create a factory for disk-based file items
            DiskFileItemFactory factory = new DiskFileItemFactory();
            factory.setSizeThreshold(1000240);

            // 2/ Create a new file upload handler
            RestletFileUpload upload = new RestletFileUpload(factory);
            List<FileItem> items;
            try {
                // 3/ Request is parsed by the handler which generates a list of FileItems
                items = upload.parseRequest(getRequest());

                Map<String, String> props = new HashMap<String, String>();
                File file = null;
                String filename = null;

                for (final Iterator<FileItem> it = items.iterator(); it.hasNext(); ) {
                    FileItem fi = it.next();
                    String name = fi.getName();
                    if (name == null) {
                        props.put(fi.getFieldName(), new String(fi.get(), "UTF-8"));
                    } else {
                        String tempDir = System.getProperty("java.io.tmpdir");
                        file = new File(tempDir + File.separator + "file.txt");
                        filename = name;
                        fi.getInputStream();
                        fi.write(file);
                    }
                }

                // [...] my processing code

                String redirectUrl = ...; // address of newly created resource
                getResponse().redirectSeeOther(redirectUrl);
            } catch (Exception e) {
                // The message of all thrown exception is sent back to
                // client as simple plain text
                getResponse().setStatus(Status.CLIENT_ERROR_BAD_REQUEST);
                e.printStackTrace();
                rep = new StringRepresentation(e.getMessage(), MediaType.TEXT_PLAIN);
            }
        } else {
            // other format != multipart form data
            getResponse().setStatus(Status.CLIENT_ERROR_BAD_REQUEST);
            rep = new StringRepresentation("Multipart/form-data required", MediaType.TEXT_PLAIN);
        }
    } else {
        // POST request with no entity.
        getResponse().setStatus(Status.CLIENT_ERROR_BAD_REQUEST);
        rep = new StringRepresentation("Error", MediaType.TEXT_PLAIN);
    }

    return rep;
}

I'll end up refactoring it to something more generic, but this is what I have by now.

回答2:

to pack it on one line, in you Restlet Resource class :

Iterator it = new RestletFileUpload(new DiskFileItemFactory()).parseRequest(getRequest()).iterator();

Then in your loop through your items, you can test whether they are or not fileItems with the method: isFormField().

Testing if a fileItem is a formField... makes sens ? ;)
but it works.

Good luck.

回答3:

To answer my own question, this is not currently feasible in version 1.2

来源：https://stackoverflow.com/questions/996819/restlet-how-to-process-multipart-form-data-requests

标签

java

restlet

multipartform-data