How to respond with HTTP 400 \"Bad Request\" in Spring MVC @ResponseBody?

Responding with HTTP 400 Error in Spring MVC @ResponseBody

When developing JSON APIs using Spring MVC's @ResponseBody approach, there may be scenarios where you need to respond with specific HTTP status codes, such as 400 "Bad Request."

Consider the following action method:

<code class="java">@RequestMapping(value = "/matches/{matchId}", produces = "application/json")
@ResponseBody
public String match(@PathVariable String matchId) {
    String json = matchService.getMatchJson(matchId);
    if (json == null) {
        // How to respond with HTTP 400 "Bad Request"?
    }
    return json;
}</code>

In this example, the method is annotated with @ResponseBody, indicating that the return value should be written directly to the HTTP response body. However, the initial code does not provide a mechanism for responding with an HTTP 400 error.

To resolve this issue, there are a couple of approaches to consider:

1. Changing the Return Type to ResponseEntity: By changing the method's return type to ResponseEntity<>, you gain access to methods that allow you to set the HTTP status code. You can then use ResponseEntity.status(HttpStatus.BAD_REQUEST).body(null) to return a 400 error response.
2. Using ResponseEntity Helper Methods: For Spring versions 4.1 and higher, ResponseEntity provides helper methods such as ResponseEntity.badRequest().body(null) and ResponseEntity.ok(json) to simplify the response creation process.

Therefore, the updated method would look like:

<code class="java">@RequestMapping(value = "/matches/{matchId}", produces = "application/json")
@ResponseBody
public ResponseEntity<String> match(@PathVariable String matchId) {
    String json = matchService.getMatchJson(matchId);
    if (json == null) {
        return ResponseEntity.badRequest().body(null);
    }
    return ResponseEntity.ok(json);
}</code>

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