Question

我有一个看起来像这样的RestController

@RestController
@RequestMapping("/user")
class UserController(val userService: UserService) {

  @PostMapping("/register")
  fun register(@RequestBody body: UsernamePasswordResource): Mono<*> {
    return userService.createUser(body.username, body.password)
        .map {ResponseEntity(it, HttpStatus.CREATED)}
        .doOnError {
          if (it is DuplicateUserException){
            ResponseEntity(ErrorResource(it.message!!), HttpStatus.BAD_REQUEST)
          } else {
            ResponseEntity(ErrorResource("Internal server error"), HttpStatus.INTERNAL_SERVER_ERROR)
          }
        }
  }
}

当我使用此端点创建使用时，会以200 OK作为响应并将其作为正文：

{
  "headers": {},
  "body": {
    "username": "test7",
    "passwordEncoded": "4HxMMUI09pltEr9pyKIxsQ==",
    "description": ""
  },
  "statusCode": "CREATED",
  "statusCodeValue": 201
}

为什么我的ResponseEntity的状态被忽略并且返回200？我该如何解决？预先感谢。

Answer 1

当终端成功返回时，默认情况下，

Spring提供 HTTP 200（确定）响应。要返回自定义的 HTTP 状态，请使用函数上方的 @ReponseStatus 批注，然后在括号中传递状态代码。您的代码段应如下所示：

@RestController
@RequestMapping("/user")
class UserController(val userService: UserService) {

  @PostMapping("/register")
  @ResponseStatus(HttpStatus.CREATED)
  fun register(@RequestBody body: UsernamePasswordResource): Mono<*> {
    return userService.createUser(body.username, body.password)
        .map {ResponseEntity(it, HttpStatus.CREATED)}
        .doOnError {
          if (it is DuplicateUserException){
            ResponseEntity(ErrorResource(it.message!!), HttpStatus.BAD_REQUEST)
          } else {
            ResponseEntity(ErrorResource("Internal server error"), HttpStatus.INTERNAL_SERVER_ERROR)
          }
        }
  }
}

Spring Webflux Mono正在忽略ResponseEntity的状态并返回200 OK

1 个答案: