我想知道如何在基于X,Y鼠标坐标的android上执行拖动吗?考虑两个简单的示例,Team Viewer / QuickSupport分别在远程智能手机和Windows Paint Pen上绘制“密码模式”。
我所能做的就是模拟触摸(使用dispatchGesture()
和也AccessibilityNodeInfo.ACTION_CLICK
)。
我找到了这些相关链接,但不知道它们是否有用:
以下是我的工作代码,该代码用于将鼠标坐标(在PictureBox
控件内部)发送到远程电话并模拟触摸。
Windows窗体应用程序:
private void pictureBox1_MouseDown(object sender, MouseEventArgs e)
{
foreach (ListViewItem item in lvConnections.SelectedItems)
{
// Remote screen resolution
string[] tokens = item.SubItems[5].Text.Split('x'); // Ex: 1080x1920
int xClick = (e.X * int.Parse(tokens[0].ToString())) / (pictureBox1.Size.Width);
int yClick = (e.Y * int.Parse(tokens[1].ToString())) / (pictureBox1.Size.Height);
Client client = (Client)item.Tag;
if (e.Button == MouseButtons.Left)
client.sock.Send(Encoding.UTF8.GetBytes("TOUCH" + xClick + "<|>" + yClick + Environment.NewLine));
}
}
编辑:
我的最后一次尝试是分别使用鼠标坐标(“ C#Windows窗体应用程序”)和自定义android例程(参考上面链接的“滑动屏幕”的代码)进行“滑动屏幕”:
private Point mdownPoint = new Point();
private void pictureBox1_MouseDown(object sender, MouseEventArgs e)
{
foreach (ListViewItem item in lvConnections.SelectedItems)
{
// Remote screen resolution
string[] tokens = item.SubItems[5].Text.Split('x'); // Ex: 1080x1920
Client client = (Client)item.Tag;
if (e.Button == MouseButtons.Left)
{
xClick = (e.X * int.Parse(tokens[0].ToString())) / (pictureBox1.Size.Width);
yClick = (e.Y * int.Parse(tokens[1].ToString())) / (pictureBox1.Size.Height);
// Saving start position:
mdownPoint.X = xClick;
mdownPoint.Y = yClick;
client.sock.Send(Encoding.UTF8.GetBytes("TOUCH" + xClick + "<|>" + yClick + Environment.NewLine));
}
}
}
private void PictureBox1_MouseMove(object sender, MouseEventArgs e)
{
foreach (ListViewItem item in lvConnections.SelectedItems)
{
// Remote screen resolution
string[] tokens = item.SubItems[5].Text.Split('x'); // Ex: 1080x1920
Client client = (Client)item.Tag;
if (e.Button == MouseButtons.Left)
{
xClick = (e.X * int.Parse(tokens[0].ToString())) / (pictureBox1.Size.Width);
yClick = (e.Y * int.Parse(tokens[1].ToString())) / (pictureBox1.Size.Height);
client.sock.Send(Encoding.UTF8.GetBytes("MOUSESWIPESCREEN" + mdownPoint.X + "<|>" + mdownPoint.Y + "<|>" + xClick + "<|>" + yClick + Environment.NewLine));
}
}
}
android AccessibilityService:
public void Swipe(int x1, int y1, int x2, int y2, int time) {
if (android.os.Build.VERSION.SDK_INT >= android.os.Build.VERSION_CODES.N) {
System.out.println(" ======= Swipe =======");
GestureDescription.Builder gestureBuilder = new GestureDescription.Builder();
Path path = new Path();
path.moveTo(x1, y1);
path.lineTo(x2, y2);
gestureBuilder.addStroke(new GestureDescription.StrokeDescription(path, 100, time));
dispatchGesture(gestureBuilder.build(), new GestureResultCallback() {
@Override
public void onCompleted(GestureDescription gestureDescription) {
System.out.println("SWIPE Gesture Completed :D");
super.onCompleted(gestureDescription);
}
}, null);
}
}
会产生以下结果(但仍然无法绘制“图案密码”,例如TeamViewer)。但是,就像下面的评论中所说的那样,我认为使用类似的方法可能可以使用“ 继续”手势来实现。欢迎向这个方向提出任何建议。
编辑2:
肯定的,解决方案是继续手势,如先前的Edit所述。
下面是我在这里找到的假定的固定代码=>
android AccessibilityService:
// Simulates an L-shaped drag path: 200 pixels right, then 200 pixels down.
Path path = new Path();
path.moveTo(200,200);
path.lineTo(400,200);
final GestureDescription.StrokeDescription sd = new GestureDescription.StrokeDescription(path, 0, 500, true);
// The starting point of the second path must match
// the ending point of the first path.
Path path2 = new Path();
path2.moveTo(400,200);
path2.lineTo(400,400);
final GestureDescription.StrokeDescription sd2 = sd.continueStroke(path2, 0, 500, false); // 0.5 second
HongBaoService.mService.dispatchGesture(new GestureDescription.Builder().addStroke(sd).build(), new AccessibilityService.GestureResultCallback(){
@Override
public void onCompleted(GestureDescription gestureDescription){
super.onCompleted(gestureDescription);
HongBaoService.mService.dispatchGesture(new GestureDescription.Builder().addStroke(sd2).build(),null,null);
}
@Override
public void onCancelled(GestureDescription gestureDescription){
super.onCancelled(gestureDescription);
}
},null);
然后,我的疑问是:如何正确发送上面代码的鼠标坐标,以哪种方式可以向任何方向执行拖动?有想法吗
编辑3:
我发现了两个用于执行拖动的例程,但是它们使用的是UiAutomation + injectInputEvent()
。AFAIK,事件注入仅在像此处和此处所述的系统应用程序中有效,我不希望如此。
这是发现的例程:
然后要实现我的目标,我认为第二个例程更适合与Edit 2上显示的代码一起使用(遵循逻辑,不包括事件注入),并分别发送pictureBox1_MouseDown
和pictureBox1_MouseMove
(C#Windows Forms Application)的所有点以Point[]
动态填充和pictureBox1_MouseUp
发送cmd执行例程并使用此数组填充。如果您对第一个例程有想法,请告诉我:D。
如果在阅读完此编辑后您有可能的解决方案,请在回答中告诉我,而我将尝试测试该想法。
pictureBox1_MouseDown
不得发送坐标。它只应存储初始坐标,然后pictureBox1_MouseUp
发送给您,因为这标志着鼠标移动的结束