java - 如何按顺序制作junit测试运行方法

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这个问题在这里已经有了答案:

How to run test methods in specific order in JUnit4? (23 个回答)

关闭 6年前.

伙计们!我有一个新问题要问你。我正在使用不同的缓存管理器将一些数据添加到缓存中，但我正面临着它的问题。我正在使用junit和spring来做。当我运行测试时，测试方法是随机执行的，但我需要它们按顺序执行。怎么做？？这是一些代码和验证控制台输出:

服务类:

@Service("HelloCache")
public class CacheServiceImpl implements CacheInterface {
    @Autowired
    @Qualifier("memcachedClient")
    private MemcachedClient mBean;

    public void Add(String key, Object object) {
        mBean.set(key, 12, object);
    }

    public void Get(String key) {
        mBean.get(key);
    }

    public void Delete(String key) {
        mBean.delete(key);
    }    
}

这里是测试:

@RunWith(SpringJUnit4ClassRunner.class)
@ContextConfiguration(locations = "file:src/main/java/spring.xml")
public class UsingMemcachedTest extends TestCase {
    @Autowired
    @Qualifier("HelloCache")
    private CacheInterface emcached;
    private byte[][] i = new byte[2500][3000];
    private String key = "j";
    @Test
    public void testAddBulkObjects() {
        System.out.println("");
        System.out.println("This is async BULK adding test");
        long time = System.currentTimeMillis();
        for (int k=1; k<=1000; k++) {
            emcached.Add(key+k, i);
        }
        long time2 = System.currentTimeMillis();
        long timeE=time2-time;
        System.out.println("Vremya add BULK objects: " + timeE);
        System.out.println("");
    }

    @Test
    public void testGetBulkObjects() {
        System.out.println("");
        System.out.println("This is getting BULK objects test");
        long time = System.currentTimeMillis(); 
        for (int k=1; k<=1000; k++) {
            emcached.Get(key+k);
        }
        long time2 = System.currentTimeMillis();
        long timeE=time2-time;
        System.out.println("Vremya Get object: " + timeE);
        System.out.println("");
    } 

    @Test
    public void testDeleteBulkObjects() {
        System.out.println("");
        System.out.println("This is deleting BULK objects test");
        long time = System.currentTimeMillis();
        for (int k=1; k<=1000; k++) {
            emcached.Delete(key+k);
        }
        long time2 = System.currentTimeMillis();
        long timeE=time2-time;
        System.out.println("Vremya delete object: " + timeE);
        System.out.println("");
    }

然后输出:

This is deleting BULK objects test
Vremya delete object: 137


This is getting BULK objects test
Vremya Get object: 703


This is async BULK adding test
Vremya add BULK objects: 87681

请帮忙!! =)

Answer 1

从 4.11 版开始，您可以使用注释指定执行顺序并按方法名称排序:

import org.junit.Test;
import org.junit.FixMethodOrder;
import org.junit.runners.MethodSorters;

@FixMethodOrder(MethodSorters.NAME_ASCENDING)
public class MyTest {

    @Test
    public void test1Create() {
        System.out.println("first");
    }

    @Test
    public void test2Update() {
        System.out.println("second");
    }
}

See JUnit 4.11 Release Notes