Espresso:如果视图存在则返回布尔值

Question

我正在尝试检查Espresso是否显示视图.这是一些伪代码来显示我正在尝试的内容:

if (!Espresso.onView(withId(R.id.someID)).check(doesNotExist()){
   // then do something
 } else {
   // do nothing, or what have you
 }

但我的问题是.check(doesNotExist())不返回布尔值.这只是一个断言.使用UiAutomator,我能够做到这样的事情:

 if (UiAutomator.getbyId(SomeId).exists()){
      .....
   }

Answer 1

测试中的条件逻辑是不合需要的.考虑到这一点,Espresso的API旨在引导测试作者远离它(通过明确的测试操作和断言).

话虽如此,您仍然可以通过实现自己的ViewAction并将isDisplayed检查(在perform方法内)捕获到AtomicBoolean来实现上述目的.

另一个不太优雅的选项 - 捕获失败检查引发的异常:

    try {
        onView(withText("my button")).check(matches(isDisplayed()));
        //view is displayed logic
    } catch (NoMatchingViewException e) {
        //view not displayed logic
    }

Answer 2

我们需要这个功能,最后我在下面实现它:

https://github.com/marcosdiez/espresso_clone

if(onView(withText("click OK to Continue")).exists()){ 
    doSomething(); 
} else { 
   doSomethingElse(); 
}

我希望它对你有用.

Answer 3

您也可以使用以下代码进行检查.如果显示视图,它将单击否则将传递.

onView(withText("OK")).withFailureHandler(new FailureHandler() {
        @Override
        public void handle(Throwable error, Matcher<View> viewMatcher){
        }
    }).check(matches(isDisplayed())).perform(customClick());

Answer 4

我认为模仿UIAutomator你可以这样做:
(.虽然,我建议重新考虑你的方法有没有条件)

ViewInteraction view = onView(withBlah(...)); // supports .inRoot(...) as well
if (exists(view)) {
     view.perform(...);
}

@CheckResult
public static boolean exists(ViewInteraction interaction) {
    try {
        interaction.perform(new ViewAction() {
            @Override public Matcher<View> getConstraints() {
                return any(View.class);
            }
            @Override public String getDescription() {
                return "check for existence";
            }
            @Override public void perform(UiController uiController, View view) {
                // no op, if this is run, then the execution will continue after .perform(...)
            }
        });
        return true;
    } catch (AmbiguousViewMatcherException ex) {
        // if there's any interaction later with the same matcher, that'll fail anyway
        return true; // we found more than one
    } catch (NoMatchingViewException ex) {
        return false;
    } catch (NoMatchingRootException ex) {
        // optional depending on what you think "exists" means
        return false;
    }
}

也exists没有分支可以实现非常简单:

onView(withBlah()).check(exists()); // the opposite of doesNotExist()

public static ViewAssertion exists() {
    return matches(anything());
}

虽然大部分时间都值得检查matches(isDisplayed()).

Answer 5

为什么没有人提到：

onView(withId(R.id.some_view_id)).check(matches(not(doesNotExist())))

not只需在 doesNotExist 之前添加即可。但如果您经常使用此逻辑，最好使用自定义匹配器。

Answer 6

基于Dhiren Mudgil的回答，我最终编写了以下方法：

public static boolean viewIsDisplayed(int viewId)
{
    final boolean[] isDisplayed = {true};
    onView(withId(viewId)).withFailureHandler(new FailureHandler()
    {
        @Override
        public void handle(Throwable error, Matcher<View> viewMatcher)
        {
            isDisplayed[0] = false;
        }
    }).check(matches(isDisplayed()));
    return isDisplayed[0];
}

我正在使用它来帮助确定当前在ViewFlipper中显示哪个视图。