如何将 Mono 错误与 Java 中的字符串进行比较 [英] How to compare Mono error to a String in Java

问题描述

我有一个类似这样的方法:

I have a method something like this :

public Mono<SomeDTO> DoAction(SomeDTO someDTOObject) {
        return findUser(someDTOObject.getUsername())
                .flatMap(existingUser -> {
                    Update update = new Update();
                   
                    return mongoTemplate.upsert(
                            Query.query(Criteria.where("username").is(someDTOObject.getUsername())),
                            update,
                            SomeDTO.class,
                            COLLECTION_NAME);                            

                }).switchIfEmpty(
                        Mono.defer(() -> {
                            return Mono.error(new Exception("User Name doesn't exist."));
                        })
                );
    }

为此，我写了一个这样的测试用例来测试异常:

For this, I have wriiten a testcase like this to test exception :

@Test
    public void DoAction_TestException() {
        SomeDTO someDTOObject = databaseUtil.SomeDTOMock;        

        Query query = Query.query(Criteria.where("username").regex("^"+userId+"$","i"));
        doReturn(Mono.empty()).when(mongoTemplate).findOne(query,
                SomeDTO.class, "COLLECTION_NAME");
        try {
            SomeDTO someDTOObjectResult = mongoImpl.DoAction(someDTOObject).block();
        }
        catch (Exception e) {
            String expected = "User Name doesn't exist.";
            String result = e.getMessage().toString();   ///////  this value during debugging is "java.lang.Exception:User Name doesn't exist. "
            assertEquals(expected,result);
        }
       
    }

当我运行上面的代码时，断言失败了，因为变量结果有额外的字符串.如何从结果中删除 java.lang.Exception ?
我不想使用任何字符串函数来删除部分字符串.任何帮助都会非常有帮助.

When I run the above code , the assert is failing becuase variable result has extra string along with it. How can I remove java.lang.Exception from the result ?
I dont want to use any string functions to remove part of string.ANy help would be very helpful.

推荐答案

Eugene 的回答已经向您展示了正确的方法.

Eugene's answer is already showing you the correct way.

替代解决方案 #1:

创建特定的异常类:EntityNotFoundException extends IOException 或 UsernameNotFoundException extends IOException 并测试您的结果是否是该类的实例.

Create a specific Exception class: EntityNotFoundException extends IOException or UsernameNotFoundException extends IOException and test if your result is an instance of that class.

替代解决方案 #2:

将您期望的字符串 expected 扩展到您真正得到的字符串.

Extend your expected String expected to the String you really get.

替代解决方案 #3:

(我不知道 Mono，所以如果 :) 如果 Mono 将异常包装到另一个异常中，您可能可以使用 e.getCause 到达原始异常:将您的行 String result = e.getMessage().toString(); 更改为 String result = e.getCause().getMessage();.

(I do not know Mono, so IF:) If Mono wraps up the Exception into another exception, you probably can reach that original exception by using e.getCause: Change your line String result = e.getMessage().toString(); to String result = e.getCause().getMessage();.

为了帮助我们找到这一点，只需在您的行 String expected = "User Name does not exist." 之后添加 e.printStackTrace(); 行.;，然后向我们展示打印了什么错误消息.从那时起，如果其他解决方案还没有帮助，我们可以进一步提供帮助.

To help us find that out, simply add the line e.printStackTrace(); right after your line String expected = "User Name doesn't exist.";, and then show us what error message was printed. From there on we can help further, if the other solutions have not helped yet.

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