在Java中如何使用JUnit验证抛出的异常？ [英] In Java how can I validate a thrown exception with JUnit?

问题描述

在为Java API编写单元测试时，您可能需要对异常执行更详细的验证。即比JUnit提供的 @test 注释提供的更多。

When writing unit tests for a Java API there may be circumstances where you want to perform more detailed validation of an exception. I.e. more than is offered by the @test annotation offered by JUnit.

例如，考虑一个应该从其他接口捕获异常的类，包装该异常并抛出包装的异常。您可能需要验证：

For example, consider an class that should catch an exception from some other Interface, wrap that exception and throw the wrapped exception. You may want to verify:

1. 确切的方法调用抛出包装的异常。




3. 包装器异常的消息。

<这里的主要内容是，您希望在单元测试中进一步验证异常（而不是关于您是否应该/ em>验证异常消息之类的争论）。

The main point here is that you want to be perf additional validation of an exception in a unit test (not a debate about whether you should verify things like the exception message).

这是一个很好的方法？

推荐答案

在JUnit 4中，可以轻松完成使用 ExpectedException 规则。

In JUnit 4 it can be easily done using ExpectedException rule.

以下是javadocs的示例：

Here is example from javadocs:

// These tests all pass.
public static class HasExpectedException {
    @Rule
    public ExpectedException thrown = ExpectedException.none();

    @Test
    public void throwsNothing() {
        // no exception expected, none thrown: passes.
    }

    @Test
    public void throwsNullPointerException() {
        thrown.expect(NullPointerException.class);
        throw new NullPointerException();
    }

    @Test
    public void throwsNullPointerExceptionWithMessage() {
        thrown.expect(NullPointerException.class);
        thrown.expectMessage("happened?");
        thrown.expectMessage(startsWith("What"));
        throw new NullPointerException("What happened?");
    }
}

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