Android可访问性：如何获得触摸的视图ID？

Question

我已经陷入困境了几天。我想确定用户触摸的观点。这是我的可访问性服务XML：

<?xml version="1.0" encoding="utf-8"?>
<accessibility-service
    android:accessibilityEventTypes="typeViewClicked"
    android:accessibilityFeedbackType="feedbackGeneric"
    android:accessibilityFlags="flagIncludeNotImportantViews|flagRetrieveInteractiveWindows|flagReportViewIds"
    android:canRetrieveWindowContent="true"
    android:description="@string/accessibility_service_description_3"
    xmlns:android="http://schemas.android.com/apk/res/android"/>

和服务：

@Override
protected void onServiceConnected() {
    super.onServiceConnected();

    AccessibilityServiceInfo configuration = new AccessibilityServiceInfo();
    configuration.eventTypes = AccessibilityEvent.TYPE_VIEW_CLICKED;
    configuration.feedbackType = AccessibilityServiceInfo.FEEDBACK_GENERIC;

    configuration.flags = AccessibilityServiceInfo.FLAG_INCLUDE_NOT_IMPORTANT_VIEWS
            | AccessibilityServiceInfo.FLAG_REPORT_VIEW_IDS
            | AccessibilityServiceInfo.FLAG_RETRIEVE_INTERACTIVE_WINDOWS;

    setServiceInfo(configuration);
}

@Override
public void onAccessibilityEvent(AccessibilityEvent event) {

    if (event.getEventType() == AccessibilityEvent.TYPE_VIEW_CLICKED) {

        //Toast to show the id of the view that was clicked
        Toast.makeText(this, "" + event.getWindowId(), Toast.LENGTH_LONG).show();

    }

}

event.getWindowid（）不起作用。我不确定它给出了谁的ID，但它为应用程序中的任何视图提供了相同的ID。我想要用户触摸的视图的ID。

Answer 1

您是否尝试过

    event.source.viewIdResourceName
    // or
    findFocusedViewInfo().viewIdResourceName

？

获取源视图ID的完全合格的资源名称。

注意：此API的主要用法用于UI测试自动化，并且为了报告客户端必须访问的源查看ID在配置AccessibilityService时，设置AccessibilityServiceInfo＃flag_report_view_ids标志。