我如何在JUnit中获取测试名称/测试状态 [英] How could i get a Testname / Test status in JUnit

问题描述

我是JUnit的新手.我正在研究获得

I am new to JUnit . I am working on the ways to get a output like

类名/函数名/状态/描述.

classname / function name / status / description.

我发现，除了使用Rule的FrameworkMethod之外，没有其他方法可以获取成功测试名称.是否可以事先在JUnitCore/textui.Thx的帮助下将规则设置为正在运行的套件！！

I figured that there is no other way to get success test name other than FrameworkMethod using Rule. Is it possible to set a Rule to a suite which s running with the help of JUnitCore/textui.Thx in advance.!!

-我在Junit 4.9b2上尝试过testwatchman，但是执行套件对我而言不起作用.任何帮助，将不胜感激.

- I tried testwatchman with Junit 4.9b2 but executing suites is not working for me. Any help would be appreciated.

@RunWith(Suite.class) 
@SuiteClasses({testclass1.class, testclass2.class}) 
public class junitCheck {
     private static String watchedLog;

     @Rule
     public MethodRule watchman= new TestWatchman() {
             @Override
             public void failed(Throwable e, FrameworkMethod method) {
                     watchedLog+= method.getName() + " " + e.getClass().getSimpleName()
                                     + "\n";
             }

             @Override
             public void succeeded(FrameworkMethod method) {
                     watchedLog+= method.getName() + " " + "success!\n";
             }
     };

}

---

public class testclass1 {

    @Test
    public void add()
    {
        Assert.assertEquals(4,2+3);
    }
}

---

public class testclass2 {


    @Test
    public void add1()
    {
        Assert.assertEquals(4,2+2);
    }
}

---

推荐答案

它不适用于该套件.因此，唯一的方法是将org.junit.rules.TestWatcher添加到测试的通用超类中.

It won't work with the suite. So the only way is to add an org.junit.rules.TestWatcher to a common super class of the tests.

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