在 Espresso 中，当多个视图匹配时如何避免 AmbiguousViewMatcherException

Question

有带有一些图像的 gridView。 gridView 的单元格来自相同的预定义布局，具有相同的 id 和 desc。

R.id.item_image == 2131493330

onView(withId(is(R.id.item_image))).perform(click());

由于网格中的所有单元格都具有相同的id，因此得到AmbiguousViewMatcherException。如何只拿起第一个或其中的任何一个？谢谢！

android.support.test.espresso.AmbiguousViewMatcherException: 'with id: is <2131493330>' matches multiple views in the hierarchy. Problem views are marked with '****MATCHES****' below.

+------------->ImageView{id=2131493330, res-name=item_image, desc=Image, visibility=VISIBLE, width=262, height=262, has-focus=false, has-focusable=false, has-window-focus=true, is-clickable=false, is-enabled=true, is-focused=false, is-focusable=false, is-layout-requested=false, is-selected=false, root-is-layout-requested=false, has-input-connection=false, x=0.0, y=0.0} ****MATCHES****

+------------->ImageView{id=2131493330, res-name=item_image, desc=Image, visibility=VISIBLE, width=262, height=262, has-focus=false, has-focusable=false, has-window-focus=true, is-clickable=false, is-enabled=true, is-focused=false, is-focusable=false, is-layout-requested=false, is-selected=false, root-is-layout-requested=false, has-input-connection=false, x=0.0, y=0.0} ****MATCHES**** |

Answer 1

你应该使用onData()来操作GridView:

onData(withId(R.id.item_image))
        .inAdapterView(withId(R.id.grid_adapter_id))
        .atPosition(0)
        .perform(click());

此代码将点击 GridView

中第一项内的图像

Answer 2

我创建了一个 ViewMatcher，它匹配它找到的第一个视图。也许这对某人有帮助。例如。当你没有 AdapterView 来使用 onData() 时。

/**
 * Created by stost on 15.05.14.
 * Matches any view. But only on first match()-call.
 */
public class FirstViewMatcher extends BaseMatcher<View> {


   public static boolean matchedBefore = false;

   public FirstViewMatcher() {
       matchedBefore = false;
   }

   @Override
   public boolean matches(Object o) {
       if (matchedBefore) {
           return false;
       } else {
           matchedBefore = true;
           return true;
       }
   }

   @Override
   public void describeTo(Description description) {
       description.appendText(" is the first view that comes along ");
   }

   @Factory
   public static <T> Matcher<View> firstView() {
       return new FirstViewMatcher();
   }
}

这样使用：

 onView(FirstViewMatcher.firstView()).perform(click());

Answer 3

与网格视图情况不完全相关，但您可以使用 hamcrest allOf 匹配器来组合多个条件：

import static org.hamcrest.CoreMatchers.allOf;

onView(allOf(withId(R.id.login_password), 
             withEffectiveVisibility(ViewMatchers.Visibility.VISIBLE)))
        .check(matches(isCompletelyDisplayed()))
        .check(matches(withHint(R.string.password_placeholder)));

Answer 4

编辑：有人在评论中提到 withParentIndex 现在可用，请先尝试一下，然后再使用下面的自定义解决方案。

令我惊讶的是，我无法通过简单地提供索引和匹配器（即 withText、withId）找到解决方案。接受的答案仅在您处理 onData 和 ListViews 时解决问题。

如果屏幕上有多个相同的视图 resId/text/contentDesc，您可以使用此自定义匹配器选择您想要的视图而不会导致 AmbiguousViewMatcherException：

public static Matcher<View> withIndex(final Matcher<View> matcher, final int index) {
    return new TypeSafeMatcher<View>() {
        int currentIndex = 0;

        @Override
        public void describeTo(Description description) {
            description.appendText("with index: ");
            description.appendValue(index);
            matcher.describeTo(description);
        }

        @Override
        public boolean matchesSafely(View view) {
            return matcher.matches(view) && currentIndex++ == index;
        }
    };
}

例如：

onView(withIndex(withId(R.id.my_view), 2)).perform(click());

将对 R.id.my_view 的第三个实例执行点击操作。

Answer 5

案例：

onView( withId( R.id.songListView ) ).perform( RealmRecyclerViewActions.scrollTo( Matchers.first(Matchers.withTextLabeled( "Love Song"))) );
onView( Matchers.first(withText( "Love Song")) ).perform( click() );

在我的Matchers.class

里面

public static Matcher<View> first(Matcher<View> expected ){

    return new TypeSafeMatcher<View>() {
        private boolean first = false;

        @Override
        protected boolean matchesSafely(View item) {

            if( expected.matches(item) && !first ){
                return first = true;
            }

            return false;
        }

        @Override
        public void describeTo(Description description) {
            description.appendText("Matcher.first( " + expected.toString() + " )" );
        }
    };
}

Answer 6

尝试了@FrostRocket 的答案，看起来最有希望，但需要添加一些定制：

public static Matcher<View> withIndex(final Matcher<View> matcher, final int index) {
    return new TypeSafeMatcher<View>() {
        int currentIndex;
        int viewObjHash;

        @SuppressLint("DefaultLocale") @Override
        public void describeTo(Description description) {
            description.appendText(String.format("with index: %d ", index));
            matcher.describeTo(description);
        }

        @Override
        public boolean matchesSafely(View view) {
            if (matcher.matches(view) && currentIndex++ == index) {
                viewObjHash = view.hashCode();
            }
            return view.hashCode() == viewObjHash;
        }
    };
}

Answer 7

您可以简单地使 NthMatcher 像：

   class NthMatcher internal constructor(private val id: Int, private val n: Int) : TypeSafeMatcher<View>(View::class.java) {
        companion object {
            var matchCount: Int = 0
        }
        init {
            var matchCount = 0
        }
        private var resources: Resources? = null
        override fun describeTo(description: Description) {
            var idDescription = Integer.toString(id)
            if (resources != null) {
                try {
                    idDescription = resources!!.getResourceName(id)
                } catch (e: Resources.NotFoundException) {
                    // No big deal, will just use the int value.
                    idDescription = String.format("%s (resource name not found)", id)
                }

            }
            description.appendText("with id: $idDescription")
        }

        public override fun matchesSafely(view: View): Boolean {
            resources = view.resources
            if (id == view.id) {
                matchCount++
                if(matchCount == n) {
                    return true
                }
            }
            return false
        }
    }

这样声明：

fun withNthId(resId: Int, n: Int) = CustomMatchers.NthMatcher(resId, n)

并像这样使用：

onView(withNthId(R.id.textview, 1)).perform(click())

Answer 8

与网格视图没有特别相关，但我有一个类似的问题，我的 RecyclerView 和 root layout 上的元素都有相同的 ID 并且两者显示到屏幕上。帮助我解决它的是检查 descendancy 例如：

 onView(allOf(withId(R.id.my_view), not(isDescendantOfA(withId(R.id.recyclerView))))).check(matches(withText("My Text")));

Answer 9

最迟 * 运行 -> 记录 Espresso 测试

通过点击不同位置的相同 ID 视图为他们生成不同的代码，所以试试吧。

它实际上解决了这些问题。

在 Espresso 中，当多个视图匹配时如何避免 AmbiguousViewMatcherException

In Espresso, how to avoid AmbiguousViewMatcherException when multiple views match

android

android-espresso