I need to receive an HTTP Post Multipart which contains only 2 parameters:
Which is the correct way to set the body? I'm going to test the HTTP call using Chrome REST console, so I'm wondering if the correct solution is to set a "label" key for the JSON parameter and the binary file.
On the server side I'm using Resteasy 2.x, and I'm going to read the Multipart body like this:
@POST
@Consumes("multipart/form-data")
public String postWithPhoto(MultipartFormDataInput multiPart) {
Map <String, List<InputPart>> params = multiPart.getFormDataMap();
String myJson = params.get("myJsonName").get(0).getBodyAsString();
InputPart imagePart = params.get("photo").get(0);
//do whatever I need to do with my json and my photo
}
Is this the way to go? Is it correct to retrieve my JSON string using the key "myJsonName" that identify that particular content-disposition? Are there any other way to receive these 2 content in one HTTP multipart request?
Thanks in advance
If I understand you correctly, you want to compose a multipart request manually from an HTTP/REST console. The multipart format is simple; a brief introduction can be found in the HTML 4.01 spec. You need to come up with a boundary, which is a string not found in the content, let’s say HereGoes. You set request header Content-Type: multipart/form-data; boundary=HereGoes. Then this should be a valid request body:
--HereGoes
Content-Disposition: form-data; name="myJsonString"
Content-Type: application/json
{"foo": "bar"}
--HereGoes
Content-Disposition: form-data; name="photo"
Content-Type: image/jpeg
Content-Transfer-Encoding: base64
<...JPEG content in base64...>
--HereGoes--
Source: Stackoverflow.com