How to process a multipart request consisting of a file and a JSON object in Spring restful service?

Question

I have the following resource (implemented using Spring 4.05.RELEASE) which accepts a file and a JSON object:

(P.S. activityTemplate is a serializable entity class)<

Answer 1

Hope this should help you. You need to set the boundary in your request to inform the HTTP Request. is simple; A brief introduction to the multipart format can be found in the below link

HTML 4.01 Specification for multipart

The following example illustrates "multipart/form-data" encoding. If the Json Object is "MyJsonObj" , and file that need to be send is "myfile.txt", the user agent might send back the following data:

Content-Type: multipart/form-data; boundary=MyBoundary

--MyBoundary
Content-Disposition: form-data; name="myJsonString"
Content-Type: application/json

MyJsonObj //Your Json Object goes here
--MyBoundary
Content-Disposition: form-data; name="files"; filename="myfile.txt"
Content-Type: text/plain

... contents of myfile.txt ...
--MyBoundary--

or if your files is of type image with name "image.gif" then,

--MyBoundary
Content-Disposition: file; filename="image.gif"
Content-Type: image/gif
Content-Transfer-Encoding: binary

...contents of image.gif...
--MyBoundary--

You specify boundary in the Content-Type header so that the server knows how to split the data sent.

So, you basically need to select a boundary value to:

• Use a value that won't appear in the HTTP data sent to the server like 'AaB03x'.
• Be consistent and use the same value all over the request.

Answer 2

The error message indicates that there is no HttpMessageConverter registered for a multi-part/MIME part of content type: application/octet-stream. Still, your jarFile parameter is most likely correctly identified as application/octet-stream, so I'm assuming there's a mismatch in the parameter mapping.

So, first try setting the same name for the parameter and the form's input element.

Another problem is that the JSON is uploaded as a (regular) value of a text input in the form, not as a separate part in the multi-part/MIME. So there's no content-type header associated with it to find out that Spring should use the JSON deserializer. You can use @RequestParam instead and register a specific converter like in this answer: JSON parameter in spring MVC controller

Answer 3

Couldn't you change your

@RequestMapping(value="/create", method=RequestMethod.POST)

to

@RequestMapping(value="/create",
                method=RequestMethod.POST, consumes ={"multipart/form-data"})