如何将API异常输出传递给自己的REST服务？ [英] How to pass API exception output to through own REST service?

问题描述

摘要：
我想通过使用我自己的Rest服务将一个REST服务端点给出的有效异常输出传递给最终用户。

Summary : I want to pass valid exception output given by one REST service end point to the end user by using my own Rest service.

什么我所做的是，我已经使用RestTemplate类在服务类中调用了该服务，它在有效的发布请求中提供了有效的输出。但是，当我向其传递无效输入时，在我调用该API的服务类中仅得到 400错误请求结果。但是，当我使用邮递员分别调用该API时，就会得到预期的输出。

What I did is, I have called that service in service class using RestTemplate class, it's giving valid output on valid post request. But when I am passing invalid input to it I am getting only '400 BAD REQUEST' result in my service class where I have called that API. But when I am calling that API separately using postman, there I'm getting expected output.

代码示例：

class Abc {
    ResponseEntity<String> = response;
    static final String url = "https://abc-xyz.com/client-rest-end-point-url";
    public ResponseEntity getDetails(RequestInput requestInput) {

        try{
            response=restTemplate.postForObject(url,requestInput,String.class);
        } catch(Exception e) {
            ResponseEntity response = (ResponseEntity<ErrorModel>)restTemplate.postForEntity(url,requestInput,ErrorModel.class);
        }//try-catch
    }//getDetails method
}//class

推荐答案

您可以为整个应用程序创建一个自定义的异常类，并且可以通过使用 JSON 发送数据。 code> throw 关键字
假设您具有异常类：

You can create a custom exception class for your entire application and you can send data in JSON by using throw keyword Suppose you have exception class is:

public class TestException extends Exception {

private static final long serialVersionUID = 1L;
private String code;
private String detailMessage;

public TestException() {
};

public TestException(String message, String code, String detailMessage) {
    super(message);
    this.code = code;
    this.detailMessage = detailMessage;
}

public TestException(String message, String code) {
    super(message);
    this.code = code;
}
//TestExceptionResponseCode is another class for message data, if required.
public TestException(TestExceptionResponseCode testExceptionResponseCode) {
    super(testExceptionResponseCode.getMessage());
    this.code = testExceptionResponseCode.getCode();
}

public String getCode() {
    return code;
}

public void setCode(String code) {
    this.code = code;
}

public String getDetailMessage() {
    return detailMessage;
}

public void setDetailMessage(String detailMessage) {
    this.detailMessage = detailMessage;
}

}

现在，在您的情况下，抛出异常可以像这样：

Now in your case throwing exception can be like :

class Abc {
ResponseEntity<String> = response;
static final String url = "https://abc-xyz.com/client-rest-end-point-url";
public ResponseEntity getDetails(RequestInput requestInput) {
       if(requestInput==null){
          throw new TestException("FAILED", "1", "Data can't be null");
    }

}

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