Posting a File and Associated Data to a RESTful WebService preferably as JSON

You can send the file and data over in one request using the multipart/form-data content type:

In many applications, it is possible for a user to be presented with a form. The user will fill out the form, including information that is typed, generated by user input, or included from files that the user has selected. When the form is filled out, the data from the form is sent from the user to the receiving application.

The definition of MultiPart/Form-Data is derived from one of those applications...

From http://www.faqs.org/rfcs/rfc2388.html:

"multipart/form-data" contains a series of parts. Each part is expected to contain a content-disposition header [RFC 2183] where the disposition type is "form-data", and where the disposition contains an (additional) parameter of "name", where the value of that parameter is the original field name in the form. For example, a part might contain a header:

Content-Disposition: form-data; name="user"

with the value corresponding to the entry of the "user" field.

You can include file information or field information within each section between boundaries. I've successfully implemented a RESTful service that required the user to submit both data and a form, and multipart/form-data worked perfectly. The service was built using Java/Spring, and the client was using C#, so unfortunately I don't have any Grails examples to give you concerning how to set up the service. You don't need to use JSON in this case since each "form-data" section provides you a place to specify the name of the parameter and its value.

The good thing about using multipart/form-data is that you're using HTTP-defined headers, so you're sticking with the REST philosophy of using existing HTTP tools to create your service.

I know that this thread is quite old, however, I am missing here one option. If you have metadata (in any format) that you want to send along with the data to upload, you can make a single multipart/related request.

The Multipart/Related media type is intended for compound objects consisting of several inter-related body parts.

You can check RFC 2387 specification for more in-depth details.

Basically each part of such a request can have content with different type and all parts are somehow related (e.g. an image and it metadata). The parts are identified by a boundary string, and the final boundary string is followed by two hyphens.

Example:

POST /upload HTTP/1.1 Host: www.hostname.com Content-Type: multipart/related; boundary=xyz Content-Length: [actual-content-length] --xyz Content-Type: application/json; charset=UTF-8 { "name": "Sample image", "desc": "...", ... } --xyz Content-Type: image/jpeg [image data] [image data] [image data] ... --foo_bar_baz-- 

Here is my approach API (i use example) - as you can see, you I don't use any file_id (uploaded file identifier to the server) in API:

1. Create photo object on server:

    POST: /projects/{project_id}/photos body: { name: "some_schema.jpg", comment: "blah"} response: photo_id 

2. Upload file (note that file is in singular form because it is only one per photo):

    POST: /projects/{project_id}/photos/{photo_id}/file body: file to upload response: - 

And then for instance:

3. Read photos list

    GET: /projects/{project_id}/photos response: [ photo, photo, photo, ... ] (array of objects) 

4. Read some photo details

    GET: /projects/{project_id}/photos/{photo_id} response: { id: 666, name: 'some_schema.jpg', comment:'blah'} (photo object) 

5. Read photo file

    GET: /projects/{project_id}/photos/{photo_id}/file response: file content 

So the conclusion is that, first you create an object (photo) by POST, and then you send second request with the file (again POST). To not have problems with CACHE in this approach we assume that we can only delete old photos and add new - no update binary photo files (because new binary file is in fact... NEW photo). However if you need to be able to update binary files and cache them, then in point 4 return also fileId and change 5 to GET: /projects/{project_id}/photos/{photo_id}/files/{fileId}.

I know this question is old, but in the last days I had searched whole web to solution this same question. I have grails REST webservices and iPhone Client that send pictures, title and description.

I don't know if my approach is the best, but is so easy and simple.

I take a picture using the UIImagePickerController and send to server the NSData using the header tags of request to send the picture's data.

NSMutableURLRequest *request = [[NSMutableURLRequest alloc] initWithURL:[NSURL URLWithString:@"myServerAddress"]]; [request setHTTPMethod:@"POST"]; [request setHTTPBody:UIImageJPEGRepresentation(picture, 0.5)]; [request setValue:@"image/jpeg" forHTTPHeaderField:@"Content-Type"]; [request setValue:@"myPhotoTitle" forHTTPHeaderField:@"Photo-Title"]; [request setValue:@"myPhotoDescription" forHTTPHeaderField:@"Photo-Description"]; NSURLResponse *response; NSError *error; [NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&error]; 

At the server side, I receive the photo using the code:

InputStream is = request.inputStream def receivedPhotoFile = (IOUtils.toByteArray(is)) def photo = new Photo() photo.photoFile = receivedPhotoFile //photoFile is a transient attribute photo.title = request.getHeader("Photo-Title") photo.description = request.getHeader("Photo-Description") photo.imageURL = "temp" if (photo.save()) { File saveLocation = grailsAttributes.getApplicationContext().getResource(File.separator + "images").getFile() saveLocation.mkdirs() File tempFile = File.createTempFile("photo", ".jpg", saveLocation) photo.imageURL = saveLocation.getName() + "/" + tempFile.getName() tempFile.append(photo.photoFile); } else { println("Error") } 

I don't know if I have problems in future, but now is working fine in production environment.

FormData Objects: Upload Files Using Ajax

XMLHttpRequest Level 2 adds support for the new FormData interface. FormData objects provide a way to easily construct a set of key/value pairs representing form fields and their values, which can then be easily sent using the XMLHttpRequest send() method.

function AjaxFileUpload() { var file = document.getElementById("files"); //var file = fileInput; var fd = new FormData(); fd.append("imageFileData", file); var xhr = new XMLHttpRequest(); xhr.open("POST", '/ws/fileUpload.do'); xhr.onreadystatechange = function () { if (xhr.readyState == 4) { alert('success'); } else if (uploadResult == 'success') alert('error'); }; xhr.send(fd); } 

https://developer.mozilla.org/en-US/docs/Web/API/FormData

Since the only missing example is the ANDROID example, I'll add it. This technique uses a custom AsyncTask that should be declared inside your Activity class.

private class UploadFile extends AsyncTask<Void, Integer, String> { @Override protected void onPreExecute() { // set a status bar or show a dialog to the user here super.onPreExecute(); } @Override protected void onProgressUpdate(Integer... progress) { // progress[0] is the current status (e.g. 10%) // here you can update the user interface with the current status } @Override protected String doInBackground(Void... params) { return uploadFile(); } private String uploadFile() { String responseString = null; HttpClient httpClient = new DefaultHttpClient(); HttpPost httpPost = new HttpPost("http://example.com/upload-file"); try { AndroidMultiPartEntity ampEntity = new AndroidMultiPartEntity( new ProgressListener() { @Override public void transferred(long num) { // this trigger the progressUpdate event publishProgress((int) ((num / (float) totalSize) * 100)); } }); File myFile = new File("/my/image/path/example.jpg"); ampEntity.addPart("fileFieldName", new FileBody(myFile)); totalSize = ampEntity.getContentLength(); httpPost.setEntity(ampEntity); // Making server call HttpResponse httpResponse = httpClient.execute(httpPost); HttpEntity httpEntity = httpResponse.getEntity(); int statusCode = httpResponse.getStatusLine().getStatusCode(); if (statusCode == 200) { responseString = EntityUtils.toString(httpEntity); } else { responseString = "Error, http status: " + statusCode; } } catch (Exception e) { responseString = e.getMessage(); } return responseString; } @Override protected void onPostExecute(String result) { // if you want update the user interface with upload result super.onPostExecute(result); } } 

So, when you want to upload your file just call:

new UploadFile().execute(); 

I wanted send some strings to backend server. I didnt use json with multipart, I have used request params.

@RequestMapping(value = "/upload", method = RequestMethod.POST) public void uploadFile(HttpServletRequest request, HttpServletResponse response, @RequestParam("uuid") String uuid, @RequestParam("type") DocType type, @RequestParam("file") MultipartFile uploadfile) 

Url would look like

http://localhost:8080/file/upload?uuid=46f073d0&type=PASSPORT 

I am passing two params (uuid and type) along with file upload. Hope this will help who don't have the complex json data to send.

You could try using https://square.github.io/okhttp/ library. You can set the request body to multipart and then add the file and json objects separately like so:

MultipartBody requestBody = new MultipartBody.Builder() .setType(MultipartBody.FORM) .addFormDataPart("uploadFile", uploadFile.getName(), okhttp3.RequestBody.create(uploadFile, MediaType.parse("image/png"))) .addFormDataPart("file metadata", json) .build(); Request request = new Request.Builder() .url("https://uploadurl.com/uploadFile") .post(requestBody) .build(); try (Response response = client.newCall(request).execute()) { if (!response.isSuccessful()) throw new IOException("Unexpected code " + response); logger.info(response.body().string()); 

@RequestMapping(value = "/uploadImageJson", method = RequestMethod.POST) public @ResponseBody Object jsongStrImage(@RequestParam(value="image") MultipartFile image, @RequestParam String jsonStr) { -- use com.fasterxml.jackson.databind.ObjectMapper convert Json String to Object }