具有以下代码：
@RequestMapping(value = "/system/login",method = RequestMethod.GET)
public void login(@RequestBody Login login) {
if(login.username == "test" && login.password == "test") {
//return HTTP 200
}
else {
//return HTTP 400
}
}
解决方法
有人在SO中提出的一种方法是抛出不同的异常,这些异常将被不同的异常处理程序所捕获：
@RequestMapping(value = "/system/login",method = RequestMethod.GET)
public void login(@RequestBody Login login) {
if(login.username == "test" && login.password == "test") {
throw new AllRightException();
}
else {
throw new AccessDeniedException();
}
}
@ExceptionHandler(AllRightException.class)
@ResponseStatus(HttpStatus.OK)
public void whenAllRight() {
}
@ExceptionHandler(AccessDeniedException.class)
@ResponseStatus(HttpStatus.BAD_REQUEST)
public void whenAccessDenied() {
}
也可以看看：@H_502_5@
> @ExceptionHandler
> @ResponseStatus@H_502_5@
BTW,你的示例代码包含错误：login.password ==“test”你应该使用equals()有:)@H_502_5@
更新：我发现另一种方法更好,因为它不使用例外：@H_502_5@
@RequestMapping(value = "/system/login",method = RequestMethod.GET)
public ResponseEntity<String> login(@RequestBody Login login) {
if(login.username == "test" && login.password == "test") {
return new ResponseEntity<String>("OK" HttpStatus.OK);
}
return new ResponseEntity<String>("ERROR",HttpStatus.BAD_REQUEST);
}
参见ResponseEntity API@H_502_5@
