TypeText无法使用

Question

尝试使用浓咖啡测试我的登录。如果我点击我的领域，那么所有的工作都很好。

ViewInteraction linearLayout = onView(
      allOf(childAtPosition(
        allOf(withId(R.id.nameEditText), 
          childAtPosition(
            withId(R.id.inputLayout), 
            0)), 
        0), 
        isDisplayed())); 
    linearLayout.check(matches(isDisplayed())) 
      .perform(click()); 

但是，如果我需要把一些文字在这个领域，我有一些问题：

ViewInteraction linearLayout = onView(
      allOf(childAtPosition(
        allOf(withId(R.id.nameEditText), 
          childAtPosition(
            withId(R.id.inputLayout), 
            0)), 
        0), 
        isDisplayed())); 
    linearLayout.check(matches(isDisplayed())) 
      .perform(typeText("SomeName")); 

并获得下一个错误：

android.support.test.espresso.PerformException: Error performing 'type text(SomeName)' on view '(Child at position 0 in parent (with id: com.fentury.android:id/nameEditText and Child at position 0 in parent with id: com.fentury.android:id/inputLayout)

OR

如果我尝试这样做：

ViewInteraction appCompatEditText = onView(
      allOf(withId(R.id.nameEditText), 
        withParent(withId(R.id.inputLayout)), isDisplayed())); 
    appCompatEditText.check(matches(isDisplayed())) 
      .perform(click()).perform(typeText("Dog")); 

我已经凌海了同样的错误：

android.support.test.espresso.PerformException: Error performing 'type text(Dog)' on view '(with id: com.fentury.android:id/nameEditText and has parent matching: with id: com.fentury.android:id/inputLayout and is displayed on the screen to the user)'

我如何解决这个错误？

Answer 1

您使用TextInputLayout吗？

下面用TextInputLayout为我工作。

onView(
      allOf(
        isDescendantOfA(withId(R.id.nameEditText)), 
        isAssignableFrom(EditText.class))) 
      .perform(typeText("some text"));