使用Espresso在索引处选择子视图

Question

使用Espresso时，使用包含子图像视图的自定义窗口小部件视图，我可以使用哪种匹配器类型来选择第n个孩子？ 例如：

+--------->NumberSlider{id=2131296844, res-name=number_slider, visibility=VISIBLE, width=700, height=95, 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=10.0, y=0.0, child-count=7}
|
+---------->NumberView{id=-1, visibility=VISIBLE, width=99, height=95, has-focus=false, has-focusable=false, has-window-focus=true, is-clickable=true, 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}
|
+---------->NumberView{id=-1, visibility=VISIBLE, width=100, height=95, has-focus=false, has-focusable=false, has-window-focus=true, is-clickable=true, 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=99.0, y=0.0}
|
+---------->NumberView{id=-1, visibility=VISIBLE, width=100, height=95, has-focus=false, has-focusable=false, has-window-focus=true, is-clickable=true, 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=199.0, y=0.0}
|
+---------->NumberView{id=-1, visibility=VISIBLE, width=100, height=95, has-focus=false, has-focusable=false, has-window-focus=true, is-clickable=true, 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=299.0, y=0.0}
|
+---------->NumberView{id=-1, visibility=VISIBLE, width=100, height=95, has-focus=false, has-focusable=false, has-window-focus=true, is-clickable=true, 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=399.0, y=0.0}
|
+---------->NumberView{id=-1, visibility=VISIBLE, width=100, height=95, has-focus=false, has-focusable=false, has-window-focus=true, is-clickable=true, 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=499.0, y=0.0}
|
+---------->NumberView{id=-1, visibility=VISIBLE, width=100, height=95, has-focus=false, has-focusable=false, has-window-focus=true, is-clickable=true, 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=599.0, y=0.0}

Answer 1

 public static Matcher<View> nthChildOf(final Matcher<View> parentMatcher, final int childPosition) {
    return new TypeSafeMatcher<View>() {
      @Override
      public void describeTo(Description description) {
        description.appendText("with "+childPosition+" child view of type parentMatcher");
      }

      @Override
      public boolean matchesSafely(View view) {
        if (!(view.getParent() instanceof ViewGroup)) {
          return parentMatcher.matches(view.getParent());
        }

        ViewGroup group = (ViewGroup) view.getParent();
        return parentMatcher.matches(view.getParent()) && group.getChildAt(childPosition).equals(view);
      }
    };
  }

使用它

onView(nthChildOf(withId(R.id.parent_container), 0)).check(matches(withText("I am the first child")));

Answer 2

为了尝试改进Maragues的解决方案，我做了一些改变。

解决方案是创建一个自定义匹配器＆lt; View＆gt; ，为父视图包装另一个 Matcher＆lt; View＆gt; ，并将子视图的索引设为匹配。

public static Matcher<View> nthChildOf(final Matcher<View> parentMatcher, final int childPosition) {
    return new TypeSafeMatcher<View>() {
        @Override
        public void describeTo(Description description) {
            description.appendText("position " + childPosition + " of parent ");
            parentMatcher.describeTo(description);
        }

        @Override
        public boolean matchesSafely(View view) {
            if (!(view.getParent() instanceof ViewGroup)) return false;
            ViewGroup parent = (ViewGroup) view.getParent();

            return parentMatcher.matches(parent)
                    && parent.getChildCount() > childPosition
                    && parent.getChildAt(childPosition).equals(view);
        }
    };
}

详细说明

您可以覆盖 describeTo 方法，以便通过附加到 Description 参数来提供对匹配器的易于理解的描述。您还需要将 describeTo 调用传播到父匹配器，以便添加说明。

@Override
public void describeTo(Description description) {
    description.appendText("position " + childPosition + " of parent "); // Add this matcher's description.
    parentMatcher.describeTo(description); // Add the parentMatcher description.
}

接下来，您应该覆盖 matchesSafely ，这将确定何时找到视图层次结构中的匹配项。当使用其父匹配提供的父匹配器的视图调用时，请检查视图是否等于提供位置处的子视图。

如果父级没有 childCount 大于子位置， getChildAt 将返回null并导致测试崩溃。最好避免崩溃并允许测试失败，以便我们获得正确的测试报告和错误消息。

@Override
public boolean matchesSafely(View view) {
if (!(view.getParent() instanceof ViewGroup)) return false; // If it's not a ViewGroup we know it doesn't match.
    ViewGroup parent = (ViewGroup) view.getParent();

    return parentMatcher.matches(parent) // Check that the parent matches.
            && parent.getChildCount() > childPosition // Make sure there's enough children.
            && parent.getChildAt(childPosition).equals(view); // Check that this is the right child.
}

Answer 3

如果您可以获得父视图。可能是this link定义了一个匹配器来获取视图的第一个孩子可能会给你一些线索。

     public static Matcher<View> firstChildOf(final Matcher<View> parentMatcher) {
        return new TypeSafeMatcher<View>() {
            @Override
            public void describeTo(Description description) {
                description.appendText("with first child view of type parentMatcher");
            }

            @Override
            public boolean matchesSafely(View view) {       

                if (!(view.getParent() instanceof ViewGroup)) {
                    return parentMatcher.matches(view.getParent());                   
                }
                ViewGroup group = (ViewGroup) view.getParent();
                return parentMatcher.matches(view.getParent()) && group.getChildAt(0).equals(view);

            }
        };
    }

Answer 4

尽管此线程中的答案确实有用，但我只是想指出，有可能无需定义新的Matcher类就可以获取特定视图的特定子句柄。

您可以通过将Espresso提供的视图匹配器合并为以下方法来实现：

/**
 * @param parentViewId the resource id of the parent [View].
 * @param position the child index of the [View] to match.
 * @return a [Matcher] that matches the child [View] which has the given [position] within the specified parent.
 */
fun withPositionInParent(parentViewId: Int, position: Int): Matcher<View> {
    return allOf(withParent(withId(parentViewId)), withParentIndex(position))
}

然后按如下所示使用此方法：

onView(
    withPositionInParent(R.id.parent, 0)
).check(
    matches(withId(R.id.child))
)