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Junit 测试中如何对异常进行断言

2022年08月03日 620Browse 0Like 0Comments

本文对在 Junit 测试中如何对异常进行断言的几种方法进行说明。

使用 Junit 5

如果你使用 Junit 5 的话,你可以直接使用 assertThrows 方法来对异常进行断言。

 

367c7ee84701784c8b3922bcc063cb73

 

代码如下:

        Exception exception = assertThrows(NumberFormatException.class, () -> {
            new Integer("one");
        });
        System.out.println(exception);

使用 AssertJ

使用 AssertJ ,你可以有不少方法进行选择。

我们尝试使用 assertThatThrownBy 和 assertThatExceptionOfType 2 个方法。

这 2 个方法的写法有点不一样,但是整体效果是差不多的。

考察如下代码:

//        AssertJ assertThatThrownBy
        assertThatThrownBy(() -> {
            new Integer("one");
        }).isInstanceOf(NumberFormatException.class).hasMessageStartingWith("For input string");

        // AssertJ assertThatExceptionOfType
        assertThatExceptionOfType(NumberFormatException.class).isThrownBy(() -> {
            new Integer("one");
        });

上面代码中,对有关断言的使用进行了一些说明。

 

https://www.ossez.com/t/junit/13989

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Last updated:2022年08月03日

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