spring-mvc File Upload Marshaling a part into an object

Example

If you want to convert the content of a part into a domain object (e.g. a User or Account or Address), then the process is very simple:

It is possible to upload multiple parts, each with a different name. For each part name, you will need one parameter annotated with @RequestPart, whose name matches the part name.

To receive a file uploaded via an HTTP Post, you need to do the following:

@RequestMapping(
    value = "...",
    method = RequestMethod.POST,
    consumes = MediaType.MULTIPART_FORM_DATA_VALUE
)
public Object uploadFile(
    @RequestPart Address address,
) {
    .
    .
    .
}

As a raw HTTP request:

POST /... HTTP/1.1
Host: ...
Content-Type: multipart/form-data; boundary=----------287032381131322

------------287032381131322
Content-Disposition: form-data; name="address"; filename="address.json"
Content-Type: application/json

{"houseNumber": "10/A", "streetName": "Dumbldore Road", "town": "Hogsmede"}
------------287032381131322--

The most important things are:

• The name of the part must match the name of the variable.
• The Content-Type of the part must be one that Spring would be able to handle if you had sent it as a regular request. That is, if you could perform a POST to an endpoint with a Content-Type of foo/bar, and Spring is able to turn that into an object, then it will also be able to marshal a part into an object.
• You must be able to set the Content-Type of the part. If you cannot, this approach will not work - Spring will not attempt to guess the Content-Type of the part.