I'm using Spring MVC for a simple JSON API, with @ResponseBody
based approach like the following. (I already have a service layer producing JSON directly.)
@RequestMapping(value = "/matches/{matchId}", produces = "application/json") @ResponseBody public String match(@PathVariable String matchId) { String json = matchService.getMatchJson(matchId); if (json == null) { // TODO: how to respond with e.g. 400 "bad request"? } return json; }
Question is, in the given scenario, what is the simplest, cleanest way to respond with a HTTP 400 error?
I did come across approaches like:
return new ResponseEntity(HttpStatus.BAD_REQUEST);
...but I can't use it here since my method's return type is String, not ResponseEntity.
The @ResponseBody annotation tells a controller that the object returned is automatically serialized into JSON and passed back into the HttpResponse object.
Sending Specific Response Status Codes The very basic way of sending response status is to use ResponseEntity object, which is returned by a controller. Controller can set a specific response status in the Response. Alternatively, we can use @ResponseStatus annotation to specify the desired status code.
@ResponseBody is a Spring annotation which binds a method return value to the web response body. It is not interpreted as a view name. It uses HTTP Message converters to convert the return value to HTTP response body, based on the content-type in the request HTTP header.
change your return type to ResponseEntity<>
, then you can use below for 400
return new ResponseEntity<>(HttpStatus.BAD_REQUEST);
and for correct request
return new ResponseEntity<>(json,HttpStatus.OK);
UPDATE 1
after spring 4.1 there are helper methods in ResponseEntity could be used as
return ResponseEntity.status(HttpStatus.BAD_REQUEST).body(null);
and
return ResponseEntity.ok(json);
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