使用 Espresso 在索引处选择子视图答案

【问题标题】：Selecting child view at index using Espresso使用 Espresso 在索引处选择子视图
【发布时间】：2014-07-15 00:44:49
【问题描述】：

当使用带有子图像视图的自定义小部件视图时，使用 Espresso 时，我可以使用哪种 Matcher 类型来选择第 n 个孩子？示例：

+--------->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}

【问题讨论】：

标签： android android-espresso

【解决方案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")));

【讨论】：

如果父级中没有足够的子视图，在最后一条指令中使用return parentMatcher.matches(view.getParent()) && view.equals(group.getChildAt(childPosition)) 会更安全，以避免出现 NullPointerException

【解决方案2】：

为了尝试改进 Maragues 的解决方案，我做了一些更改。

解决方案是创建一个自定义的Matcher，它为父视图包装另一个Matcher，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 参数来提供易于理解的匹配器描述。您还需要将 describeTo 调用传播到父匹配器，这样它的描述也会被添加。

@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，您应该重写 matchesSafely，这将确定何时在视图层次结构中找到匹配项。当使用其父级匹配提供的父级匹配器的视图调用时，请检查该视图是否等于提供位置的子级。

如果父级没有大于子级位置的 childCount，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.
}

【讨论】：

很好的答案，感谢您非常详细的解释。

【解决方案3】：

如果你能得到父视图。可能是this link，它定义了一个匹配器来获取视图的第一个孩子可以给你一些线索。

     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);

            }
        };
    }

【讨论】：

如果需要获取第n个，应该使用getChildAt(n)。
它对我有用。谢谢。我建议改进以重用该解决方案，使用名为 isChildOfAtPosition(@IdRes parentId: Int, position: Int) gist.github.com/luisfernandezbr/… 的匹配器

【解决方案4】：

虽然此线程中的答案确实有效，但我只是想指出，无需定义新的 Matcher 类即可获得特定视图的特定子级的句柄。

您可以通过将 Espresso 提供的视图匹配器加入如下方法来实现：

/**
 * @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))
}

然后使用这个方法如下：

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

【讨论】：