Question

你能不能告诉我，通常的做法是编写一个抛出异常的JUnit Test方法，例如

class A {
    public String f(int param) throws Exception {
        if (param == 100500)
            throw new Exception();
        return "";
    }
}

private A object = new A();

@Test
public void testSomething() throws Exception {
    String expected = "";
    assertEquals(object.f(5), expected);
}

实际上，方法f()不会为该参数抛出异常（5）但是我必须声明该异常。

Answer 1

是的，它完全正常，如果它确实抛出异常，则测试将被视为失败。

您需要指定该方法抛出Exception，即使您知道具体情况没有（此检查由编译器完成）。

在这种情况下，您期望的是object.f(5)返回一个空字符串。任何其他结果（非空字符串或抛出异常）都会导致测试用例失败。

Answer 2

如果您正在调用的方法抛出一个已检查的异常，则您需要尝试捕获或重新抛出。从测试本身做到这一点很好。有多种方法可以使用JUnit测试Exception。我试图在下面提供一个简短的摘要：

import org.junit.Rule;
import org.junit.Test;
import org.junit.rules.ExpectedException;

/**
 * Example uses Kent Beck - Test Driven Development style test naming
 * conventions
 */
public class StackOverflowExample {

    @Rule
    public ExpectedException expectedException = ExpectedException.none();

    @Test
    // Note the checked exception makes us re-throw or try / catch (we're
    // re-throwing in this case)
    public void calling_a_method_which_throws_a_checked_exception_which_wont_be_thrown() throws Exception {
        throwCheckedException(false);
    }

    /*
     * Put the class of the specific Exception you're looking to trigger in the
     * annotation below. Note the test would fail if it weren't for the expected
     * annotation.
     */
    @Test(expected = Exception.class)
    public void calling_a_method_which_throws_a_checked_exception_which_will_be_thrown_and_asserting_the_type()
            throws Exception {
        throwCheckedException(true);
    }

    /*
     * Using ExpectedException we can also test for the message. This is my
     * preferred method.
     */
    @Test
    public void calling_a_method_which_throws_a_checked_exception_which_will_be_thrown_and_asserting_the_type_and_message()
            throws Exception {
        expectedException.expect(Exception.class);
        expectedException.expectMessage("Stack overflow example: checkedExceptionThrower");
        throwCheckedException(true);
    }

    // Note we don't need to rethrow, or try / catch as the Exception is
    // unchecked.
    @Test
    public void calling_a_method_which_throws_an_unchecked_exception() {
        expectedException.expect(Exception.class);
        expectedException.expectMessage("Stack overflow example: uncheckedExceptionThrower");
        throwUncheckedException();
    }

    private void throwCheckedException(boolean willThrow) throws Exception {
        // Exception is a checked Exception
        if (willThrow) {
            throw new Exception("Stack overflow example: checkedExceptionThrower");
        }
    }

    private void throwUncheckedException() throws NullPointerException {
        // NullPointerException is an unchecked Exception
        throw new NullPointerException("Stack overflow example: uncheckedExceptionThrower");
    }

}

Answer 3

JUnit-Test旨在测试给定方法的正确行为。测试方法抛出错误（例如，错误的参数）是一个完全有效的方案。如果它是一个经过检查的异常，您必须将其添加到测试方法声明中或在方法中捕获它并将Assert设置为false（如果不应该发生异常）。

您可以使用expected注释中的@Test字段告诉JUnit，如果发生异常，此测试应该通过。

@Test(expected = Exception.class)
public void testSomething() throws Exception {
    String expected = "";
    assertEquals(object.f(5), expected);
}

在这种情况下，测试方法应抛出异常，因此测试将通过。如果从注释中删除expected = Exception.class，则在发生异常时测试将失败。

Answer 4

您可以使用以下方法测试启动异常：

@Test(expected = ValidationException.class)
public void testGreaterEqual() throws ValidationException {
    Validate.greaterEqual(new Float(-5), 0f, "error7");
}

junit测试方法可以抛出异常吗？

4 个答案: