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无法在Android JUnit 4检测测试中启动未绑定服务

android-testing android

Raman Budny · 技术社区 · 10 年前

我正在尝试使用测试支持库框架测试基本的未绑定服务: http://developer.android.com/tools/testing-support-library/index.html

本地服务.java:

public class LocalService extends Service {
@Override
public IBinder onBind(Intent intent) {
        return null;
    }
}

本地服务测试.java:

@RunWith(AndroidJUnit4.class)
public class LocalServiceTest {
    @Rule
    public final ServiceTestRule mServiceRule = new ServiceTestRule();

    @Test
    public void testWithUnboundService() throws TimeoutException {
        Intent serviceIntent =
            new Intent(InstrumentationRegistry.getTargetContext(), LocalService.class);

        mServiceRule.startService(serviceIntent);
    }
}

当我尝试运行测试时,我得到以下TimeoutException:

java.util.concurrent.TimeoutException: Waited for 5 SECONDS, but service was never connected
at android.support.test.rule.ServiceTestRule.waitOnLatch(ServiceTestRule.java:258)
at android.support.test.rule.ServiceTestRule.bindServiceAndWait(ServiceTestRule.java:207)
at android.support.test.rule.ServiceTestRule.startService(ServiceTestRule.java:137)
at com.example.android.testing.ServiceTestRuleSample.LocalServiceTest.testWithBoundService(LocalServiceTest.java:71)
at java.lang.reflect.Method.invokeNative(Native Method)
at java.lang.reflect.Method.invoke(Method.java:515)
at org.junit.runners.model.FrameworkMethod$1.runReflectiveCall(FrameworkMethod.java:50)
at org.junit.internal.runners.model.ReflectiveCallable.run(ReflectiveCallable.java:12)
at org.junit.runners.model.FrameworkMethod.invokeExplosively(FrameworkMethod.java:47)
at org.junit.internal.runners.statements.InvokeMethod.evaluate(InvokeMethod.java:17)
at android.support.test.rule.ServiceTestRule$ServiceStatement.evaluate(ServiceTestRule.java:329)
at org.junit.rules.RunRules.evaluate(RunRules.java:20)
at org.junit.runners.ParentRunner.runLeaf(ParentRunner.java:325)
at org.junit.runners.BlockJUnit4ClassRunner.runChild(BlockJUnit4ClassRunner.java:78)
at org.junit.runners.BlockJUnit4ClassRunner.runChild(BlockJUnit4ClassRunner.java:57)
at org.junit.runners.ParentRunner$3.run(ParentRunner.java:290)
at org.junit.runners.ParentRunner$1.schedule(ParentRunner.java:71)
at org.junit.runners.ParentRunner.runChildren(ParentRunner.java:288)
at org.junit.runners.ParentRunner.access$000(ParentRunner.java:58)
at org.junit.runners.ParentRunner$2.evaluate(ParentRunner.java:268)
at org.junit.runners.ParentRunner.run(ParentRunner.java:363)
at org.junit.runners.Suite.runChild(Suite.java:128)
at org.junit.runners.Suite.runChild(Suite.java:27)
at org.junit.runners.ParentRunner$3.run(ParentRunner.java:290)
at org.junit.runners.ParentRunner$1.schedule(ParentRunner.java:71)
at org.junit.runners.ParentRunner.runChildren(ParentRunner.java:288)
at org.junit.runners.ParentRunner.access$000(ParentRunner.java:58)
at org.junit.runners.ParentRunner$2.evaluate(ParentRunner.java:268)
at org.junit.runners.ParentRunner.run(ParentRunner.java:363)
at org.junit.runner.JUnitCore.run(JUnitCore.java:137)
at org.junit.runner.JUnitCore.run(JUnitCore.java:115)
at android.support.test.internal.runner.TestExecutor.execute(TestExecutor.java:54)
at android.support.test.runner.AndroidJUnitRunner.onStart(AndroidJUnitRunner.java:240)
at android.app.Instrumentation$InstrumentationThread.run(Instrumentation.java:1729)

有可能用ServiceTestRule测试此类服务吗? 如果没有,是否有其他合适的方法使用JUnit4测试类启动和停止服务?

3 回复 | 直到 10 年前

Huiyang Shan 8 年前

我遇到了和你一样的问题。当我在Android参考中搜索原因时 ServiceTestRule ,我发现原因很简单。

注意:此方法仅支持允许客户端绑定到的服务他们换句话说,如果您的服务onBind(Intent)方法返回 null,则将引发TimeoutException。

因此,我更改了onBind()方法以返回非空绑定器。

@Override
public IBinder onBind(Intent i) {
    return new LocalBinder();
}

其中LocalBinder定义为

public class LocalBinder extends Binder {
    LocalService getService() {
        return LocalService.this;
    }
}

然后TimeoutException消失了!

我想测试失败的原因是您的测试运行程序无法绑定正在测试的服务。 尽管在您的案例中启动服务后,服务正在运行,但测试运行程序无法检测服务是否正在运行。 它必须将测试用例标记为失败。

Saeed Entezari 6 年前

试试这个

@Rule
public final ServiceTestRule mServiceRule = ServiceTestRule.withTimeout(60L, TimeUnit.SECONDS);

EnduroDave 10 年前

这对我有用

@Test(timeout=1000 * 60)
public void testWithStartedService(){
    try {
        mServiceRule.startService(
                new Intent(InstrumentationRegistry.getTargetContext(), UpdaterService.class));
    } catch (TimeoutException e) {
        e.printStackTrace();
    }
    //do something
}