Question

我正在尝试使用Process.Start()通过C＃打开一个应用程序，然后在应用程序上模拟鼠标点击。

当我运行应用程序时，左上角会出现鼠标点击，这不是我想要的。

public partial class Form1 : Form
{
    [DllImport("user32.dll", CharSet = CharSet.Auto,
                             CallingConvention = CallingConvention.StdCall)]
    public static extern void mouse_event(uint dwFlags, int dx, int dy,
                                          int dwData, int dwExtraInfo);
    [DllImport("user32.dll")]
    public static extern bool SetForegroundWindow(IntPtr hWnd);

    private const int MOUSEEVENTF_ABSOLUTE = 0x8000;
    private const int MOUSEEVENTF_LEFTDOWN = 0x0002;
    private const int MOUSEEVENTF_LEFTUP = 0x0004;
    private const int MOUSEEVENTF_MIDDLEDOWN = 0x0020;
    private const int MOUSEEVENTF_MIDDLEUP = 0x0040;
    private const int MOUSEEVENTF_MOVE = 0x0001;
    private const int MOUSEEVENTF_RIGHTDOWN = 0x0008;
    private const int MOUSEEVENTF_RIGHTUP = 0x0010;
    private const int MOUSEEVENTF_WHEEL = 0x0800;
    private const int MOUSEEVENTF_XDOWN = 0x0080;
    private const int MOUSEEVENTF_XUP = 0x0100;

    private void button1_Click(object sender, EventArgs e)
    {
        new Thread(() =>
        {
            AddTextToListBox1(label1);            
        }).Start();
        new Thread(() =>
        {
           // AddTextToListBox2(label2);
        }).Start();
    }

    private void AddTextToListBox1(Label label)
    {
        if (label1.InvokeRequired)
        {
            stringDelegate sd = new stringDelegate(AddTextToListBox1);
            this.Invoke(sd, new object[] { label1 });
        }
        else
        {
            label1.Text = this.ToString()+"Hi";
            System.Diagnostics.Process p = new System.Diagnostics.Process();
            p.StartInfo.FileName = "C://xyz//xyz.exe";
            p.Start();
            IntPtr hwnd = p.MainWindowHandle;
        }
    }

    private void AddTextToListBox2(Label label)
    {
        if (label1.InvokeRequired)
        {
            stringDelegate sd = new stringDelegate(AddTextToListBox2);
            this.Invoke(sd, new object[] { label1 });
        }
        else
        {
            label2.Text = "done";
            Thread.Sleep(6000);
            int X = 579;
            int Y = 637;
            mouse_event(MOUSEEVENTF_ABSOLUTE| MOUSEEVENTF_MOVE, X, Y, 0, 0);
            mouse_event(MOUSEEVENTF_LEFTDOWN | MOUSEEVENTF_LEFTUP, X, Y, 0, 0);              
        }
    }

Answer 1

我认为mouse_event只会在当前应用中引发事件，因为它不会在params上使用hwnd。我在这方面的经验很少，但我做了一些研究。看看SendMessage函数。

MSDN article：

将指定的消息发送到一个或多个窗口。 SendMessage函数调用指定窗口的窗口过程并且在窗口过程处理完消息之前不会返回。

您可以使用它发送许多消息，包括鼠标事件

如何在已启动的进程上模拟鼠标单击

1 个答案: