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

时间:2018-08-26 01:35:42

标签: java android rect accessibilityservice

我正在尝试使用以下代码执行点击,但是here表示存在一些限制,我可以在测试中看到这一点。

但是似乎该错误是由于Rect()不包含 X Y 导致的,为什么每次我在不支持的位置单击(可能在示例findSmallestNodeAtPoint()的这一行执行了一些限制,例如在示例的自述文件上说的那样:

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

Here在这个问题中,还提到了Github的这个示例,并给出了一个代码示例的答案,该示例工作自android Nougat +(API 24及更高版本),且100%(测试),但我需要将以下代码修复到我的应用中,还支持在早期版本的android中执行点击。

然后已经知道此代码在哪里失败了,我想知道与以下内容相关的内容:

  

!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();
}

1 个答案:

答案 0 :(得分:3)

好的,我已经回答了我自己的问题,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,似乎无法逃避基于this的使用Rect()getBoundsInScreen()的使用。

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。但是比上面的问题链接的代码更好:D。

在我的测试中,仅无法在android虚拟键盘上执行点击。