如何在java中捕获400个错误请求

我必须捕获400错误请求并相应地执行操作

我有一个解决方案，它按预期工作：

try {

//some rest api code
} catch (HttpStatusCodeException e) {
            if (e.getStatusCode() == HttpStatus.BAD_REQUEST) {
                // Handle Bad Request and perform operation according to that..

            }
}

但我不知道捕获HttpStatusCodeException然后检查状态代码是否是一种好方法

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有人能提出另一种处理400个错误请求的方法吗？

就是这样，用ControllerAdvice声明GlobalExceptionHandler

@ControllerAdvice
public class GlobalExceptionHandler {

@ExceptionHandler(KeyNotFoundException.class)
    public ResponseEntity<ExceptionResponse> keyNotFound(KeyNotFoundException ex) {
        ExceptionResponse response = new ExceptionResponse();
        response.setStatusCode(HttpStatus.BAD_REQUEST.toString().substring(0,3));
        response.setError(HttpStatus.BAD_REQUEST.getReasonPhrase());
        response.setMessage(ex.getMessage());
        response.setTimestamp(LocalDateTime.now());
        return new ResponseEntity<ExceptionResponse>(response, HttpStatus.BAD_REQUEST);
    }
}

例外响应

public class ExceptionResponse {
    
    private String error;
    private String message;
    private String statusCode;
    @JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "dd-MM-yyyy hh:mm:ss")
    private LocalDateTime timestamp;
    
// getters and setters

在api代码捕获块中，您发现请求中缺少参数或密钥

throw new  KeyNotFoundException ("Key not found in the request..."+ex.getMessage());

---

HttpStatusCodeException

是一个抽象类

HttpClientErrorException.BadRequest

是一个可以直接捕获的具体异常。

throw new  KeyNotFoundException ("Key not found in the request..."+ex.getMessage());