Question

我想知道如何在抛出异常时从rest端点返回自定义状态代码和消息。下面的代码允许我在抛出UserDuplicatedException时抛出我自己的自定义状态代码571，但我似乎无法找到给它一个额外的消息或错误原因的方法。你能帮忙吗？

@ControllerAdvice
public class ExceptionResolver {

@ExceptionHandler(UserDuplicatedException.class)
public void resolveAndWriteException(Exception exception, HttpServletResponse response) throws IOException {
    int status = 571;
    response.setStatus(status);
}

}

Answer 1

这应该是直截了当的。

创建自定义错误类：

public class Error {
    private String statusCode;
    private String message;
    private List<String> errors;
    private Date date;

    public Error(String status, String message) {
        this.statusCode = status;
        this.message = message;
    }

    //Getters - Setters 
}

在您的@ControllerAdvice中

@ControllerAdvice
public class ExceptionResolver {

    @ExceptionHandler(UserDuplicatedException.class)
    public ResponseEntity<Error> resolveAndWriteException(UserDuplicatedException e) throws IOException {
        Error error = new Error("571", e.getMessage());
        error.setErrors(//get your list or errors here...);
        return new ResponseEntity<Error>(error, HttpStatus.Select-Appropriate); 
    }
}

Answer 2

你可以在响应中返回一个json，你可以使用Gson将一个对象转换为JSON，请试试这个：

public class Response{
   private Integer status;
   private String message;

   public String getMessage(){
      return message;
   }

   public void setMessage(String message){
     this.message = message;
   }

   public Integer getStatus(){
      return status;
   }

   public Integer setStatus(Integer status){
      this.status = status;
   }
}    

@ExceptionHandler(UserDuplicatedException.class)
@ResponseBody
public String resolveAndWriteException(Exception exception) throws IOException {
    Response response = new Response();
    int status = 571;
    response.setStatus(571);
    response.setMessage("Set here the additional message");
    Gson gson = new Gson();
    return gson.toJson(response);
}

在Spring中向@ExceptionHandler添加自定义消息

2 个答案: