getRootInActiveWindow 中的 AccessibilityNodeInfo:执行单击失败,Rect() 不包含 X、Y 坐标

AccessibilityNodeInfo in getRootInActiveWindow: Perform click fails with Rect() not contains X, Y coordenates

我正在尝试使用以下代码执行点击,但 here 说存在一些限制,我可以在我的测试中看到这一点。

但是这个bug好像是因为Rect()不包含XY坐标,为什么每次点击在不支持的地方(可能有一些限制,就像在示例的自述文件中所说的那样)在 findSmallestNodeAtPoint() 例程中执行此行:

if (!bounds.contains(x, y)) {
      System.out.println("ERROR DETECTED!!! :::::: NOT bounds.contains(x, y) :::::::");
      return null;
   }

在这个问题中,也提到了这个 Github 的例子,并给出了一个代码示例的答案,该代码示例 100%(测试)来自 android Nougat+(api 24 及以上),但在我的情况下,我需要将下面的代码修复到我的应用程序中,还支持在以前版本的 android.

中执行点击

那么已经知道这段代码哪里出错了,我想知道可以做些什么:

!bounds.contains(x, y)

代码的主要部分是:

private static void logNodeHierachy(AccessibilityNodeInfo nodeInfo, int depth) {
    Rect bounds = new Rect();
    nodeInfo.getBoundsInScreen(bounds);

    StringBuilder sb = new StringBuilder();
    if (depth > 0) {
        for (int i=0; i<depth; i++) {
            sb.append("  ");
        }
        sb.append("\u2514 ");
    }
    sb.append(nodeInfo.getClassName());
    sb.append(" (" + nodeInfo.getChildCount() +  ")");
    sb.append(" " + bounds.toString());
    if (nodeInfo.getText() != null) {
        sb.append(" - \"" + nodeInfo.getText() + "\"");
    }
    System.out.println(sb.toString());

    for (int i=0; i<nodeInfo.getChildCount(); i++) {
        AccessibilityNodeInfo childNode = nodeInfo.getChild(i);
        if (childNode != null) {
            logNodeHierachy(childNode, depth + 1);
        }
    }
}

private static AccessibilityNodeInfo findSmallestNodeAtPoint(AccessibilityNodeInfo sourceNode, int x, int y) {
    Rect bounds = new Rect();
    sourceNode.getBoundsInScreen(bounds);

    if (!bounds.contains(x, y)) {
        System.out.println(":::::: NOT bounds.contains(x, y) :::::::");
        return null;
    }

    for (int i=0; i<sourceNode.getChildCount(); i++) {
        AccessibilityNodeInfo nearestSmaller = findSmallestNodeAtPoint(sourceNode.getChild(i), x, y);
        if (nearestSmaller != null) {
            return nearestSmaller;
        }
    }
    return sourceNode;
}

public void click(int x, int y) {
    System.out.println(String.format("Click [%d, %d]", x, y));
    AccessibilityNodeInfo nodeInfo = getRootInActiveWindow();
    if (nodeInfo == null) return;
    AccessibilityNodeInfo nearestNodeToMouse = findSmallestNodeAtPoint(nodeInfo, x, y);
    if (nearestNodeToMouse != null) {
        logNodeHierachy(nearestNodeToMouse, 0);
        nearestNodeToMouse.performAction(AccessibilityNodeInfo.ACTION_CLICK);
    }
    nodeInfo.recycle();
}

好吧,我回答我自己的问题,AccessibilityService 的专家已经认为我正在构建恶意软件而不喜欢帮助这类人。


显然似乎不可能逃避使用 Rect()getBoundsInScreen() 基于 this 答案。

我还发现了另一个类似的代码,ref:

public void click(int x, int y) {

    clickAtPosition(x, y, getRootInActiveWindow());
}

public static void clickAtPosition(int x, int y, AccessibilityNodeInfo node) {
    if (node == null) return;

    if (node.getChildCount() == 0) {
        Rect buttonRect = new Rect();
        node.getBoundsInScreen(buttonRect);
        if (buttonRect.contains(x, y)) {
            // Maybe we need to think if a large view covers item?
            node.performAction(AccessibilityNodeInfo.ACTION_CLICK);
            System.out.println("1º - Node Information: " + node.toString());
        }
    } else {
        Rect buttonRect = new Rect();
        node.getBoundsInScreen(buttonRect);
        if (buttonRect.contains(x, y)) {
            // Maybe we need to think if a large view covers item?
            node.performAction(AccessibilityNodeInfo.ACTION_CLICK);
            System.out.println("2º - Node Information: " + node.toString());
        }
        for (int i = 0; i < node.getChildCount(); i++) {
            clickAtPosition(x, y, node.getChild(i));
        }
    }
}

也有麻烦。但是比上面问题链接的代码更好:D。

在我的测试中,只有无法执行单击 android 虚拟键盘。