代码之家 › 专栏 › 技术社区 › Shashank

spring boot返回400缺少构造函数no stacktrace

spring-boot spring-mvc spring java

Shashank · 技术社区 · 6 年前

我接到一个电话

public Test postTest(@RequestBody Test test) {
}


public class Test {
    @JsonProperty("EMAIL_ADDRESS")
    @NotNull(message = "EMAIL_ADDRESS_NOT_NULL")
    @Size(min = 1, message = "EMAIL_ADDRESS_NOT_EMPTY")
    @Email(flags = Flag.CASE_INSENSITIVE, message = "EMAIL_INVALID")
    private String emailAddress;

    public ForgetPassword(
            @NotNull(message = "EMAIL_ADDRESS_NOT_NULL") @Size(min = 1, 
            message = "EMAIL_ADDRESS_NOT_EMPTY") @Email(flags = 
            Flag.CASE_INSENSITIVE, 
            message = "EMAIL_INVALID") String emailAddress) {
        super();
        this.emailAddress = emailAddress;
    }

    public String getEmailAddress() {
        return emailAddress;
    }

    public void setEmailAddress(String emailAddress) {
        this.emailAddress = emailAddress;
    }
}

我错过了默认的构造器,当我试图发帖时,它一直在返回错误的请求。如果我添加默认构造函数,它就会工作。它甚至不会抛出堆栈跟踪。我正在使用自定义模板发送回中的响应。我正在重写badrequestexception以发回自定义消息。

@ResponseStatus(HttpStatus.BAD_REQUEST)
public class BadRequestException extends RuntimeException {
    public BadRequestException(String message) {
        super(message);
    }
}

有什么方法我应该重写来捕获这些异常。我想知道为什么出了问题,并且能够看到堆栈的错误。

2 回复 | 直到 6 年前

Gaurav Srivastav 6 年前

您可以使用 @ControllerAdvice . 您可以使用 @ExcecptionHandler . 您可以记录错误跟踪以进行调试。

package com.startwithjava.exception;
import org.springframework.http.HttpStatus;
import org.springframework.http.ResponseEntity;
import org.springframework.web.bind.MethodArgumentNotValidException;
import org.springframework.web.bind.annotation.ControllerAdvice;
import org.springframework.web.bind.annotation.ExceptionHandler;

@ControllerAdvice
public class ApiExceptionHandler {
    @ExceptionHandler(Exception.class)
    public ResponseEntity<Object> handlerGenericError(Exception ex){
        ex.printStackTrace();
        return new ResponseEntity<>(ex.getMessage(), HttpStatus.INTERNAL_SERVER_ERROR);
    }
    @ExceptionHandler(BadRequestException.class)
    public ResponseEntity<Object> handlerBadRequest(BadRequestException ex){
        return new ResponseEntity<>(ex.getMessage(), HttpStatus.NOT_FOUND);
    }
}

Alien 6 年前

您可以像下面这样声明一个类来处理异常。

import org.slf4j.Logger;
import org.slf4j.LoggerFactory;
import org.springframework.http.HttpStatus;
import org.springframework.web.bind.annotation.ControllerAdvice;
import org.springframework.web.bind.annotation.ExceptionHandler;
import org.springframework.web.bind.annotation.ResponseStatus;

@ControllerAdvice
public class GlobalExceptionHandler {
    private static Logger LOGGER = LoggerFactory.getLogger(GlobalExceptionHandler.class);

    @ExceptionHandler(Exception.class)
    @ResponseStatus(value=HttpStatus.INTERNAL_SERVER_ERROR, reason="Error processing the request, please check the logs")
    public void handleException(Exception e) {
        LOGGER.error("Exception occurred", e);
    }
}