Mockito 如何模拟和断言抛出的异常? [英] Mockito How to mock and assert a thrown exception?

问题描述

我在 junit 测试中使用 mockito.您如何使异常发生然后断言它具有(通用伪代码)

I'm using mockito in a junit test. How do you make an exception happen and then assert that it has (generic pseudo-code)

推荐答案

BDD Style解决方案(更新到 Java 8)

Mockito 单独不是处理异常的最佳解决方案，使用 Mockito 和 Catch-Exception

BDD Style Solution (Updated to Java 8)

Mockito alone is not the best solution for handling exceptions, use Mockito with Catch-Exception

given(otherServiceMock.bar()).willThrow(new MyException());

when(() -> myService.foo());

then(caughtException()).isInstanceOf(MyException.class);

示例代码

• Mockito + Catch-Exception + Assertj 完整示例
• eu.codearte.catch-exception:捕获异常:2.0
• org.assertj:assertj-core:3.12.2

这篇关于Mockito 如何模拟和断言抛出的异常?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！