java - Spring项目中如何处理异常？

Question

这个问题已经有答案了:

Where to handle Exceptions in Spring Applications (6 个回答)

已关闭 3 年前。

正在写我的第一次战斗Spring-MVC项目。客户类(class)中有一个钱包(double钱包字段)。客户有机会从账户中提取资金。如果用户尝试提取的资金多于帐户可用的资金，我想在该方法中抛出错误。如何做到这一点？

我怎样才能摆脱if else构造并处理没有它们的方法 try catch 。或者用 if 更好？

客户服务:

package com.tinychiefdelights.service;

import com.tinychiefdelights.model.Customer;
import com.tinychiefdelights.repository.CustomerRepository;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.stereotype.Service;

@Service
public class CustomerService extends UserService {


    private CustomerRepository customerRepository;


    @Autowired
    public void setCustomerRepository(CustomerRepository customerRepository) {
        this.customerRepository = customerRepository;
    }



    // Методы
    //

    public void withdrawMoney(Customer customer, double money){ 
        if (money <= customer.getWallet()) { // We make sure that the amount indicated by the customer is less than the wallet
            customer.setWallet(customer.getWallet() - money);
        } else {
            //////////!!!! OR TRY CATCH and how to implement?!!!!!!!!
        }
    }

}

Answer 1

创建您的自定义异常类，例如OverWithDrawException注册一个 Spring Controller 来处理带有 HTTP 状态码的异常

@Controller
public class ExceptionHandlingController {

  // @RequestHandler methods
  ...

  // Exception handling methods

  // Convert a predefined exception to an HTTP Status code
  @ResponseStatus(value=HttpStatus.CONFLICT,
                  reason="Account over with draw attempt")  // 409
  @ExceptionHandler(OverWithDrawException.class)
  public void overWithDrawHandle() {
    // Nothing to do
  }

@ExceptionHandler(Exception.class)
  public ModelAndView handleError(HttpServletRequest req, Exception ex) {
    logger.error("Request: " + req.getRequestURL() + " raised " + ex);

    ModelAndView mav = new ModelAndView();
    mav.addObject("exception", ex);
    mav.addObject("url", req.getRequestURL());
    mav.setViewName("error");
    return mav;
  }

// Specify name of a specific view that will be used to display the error:
  @ExceptionHandler({SQLException.class,DataAccessException.class})
  public String databaseError() {
    // Nothing to do.  Returns the logical view name of an error page, passed
    // to the view-resolver(s) in usual way.
    // Note that the exception is NOT available to this view (it is not added
    // to the model) but see "Extending ExceptionHandlerExceptionResolver"
    // below.
    return "databaseError";
  }

}

从 https://spring.io/blog/2013/11/01/exception-handling-in-spring-mvc 复制上述示例我很确定会有更新更好的方法来做到这一点。