send mouse click to winamp

Question

I'm trying to learn mouse/keyboard emulation using win api.
I found out that it is possible to emulate button click using sendmessage() function.

Well, I get coordinates of "play" button (it is (60;100)) and trying to push this button using the following code:

            int x = 60;
            int y = 100;
            int lParam = ((x << 16) | (y & 0xffff));

            int parentWindow = FindWindow("BaseWindow_RootWnd", "Main Window");//get main winamp window
            MessageBox.Show(parentWindow.ToString());//failed if 0

            SendMessage(parentWindow, WM_LBUTTONDOWN, IntPtr.Zero, new IntPtr(lParam));//send left mouse button down
            SendMessage(parentWindow, WM_LBUTTONUP, IntPtr.Zero, new IntPtr(lParam));//send left mouse button up

But this code has no effect on the winamp.

Can anybody point me on the mistakes i made? Any help is greatly appreciated!

ps It's not applicable for me to move mouse to the winamp play button, than do a click, than move it back.
Also for winamp it is impossible to get button's handle. With button handle SendMessage() works fairly well, but with coordinates it doesn't at all.

ADDITION #1

Well, the code above activates winamp window and shows it, if it was minimized. But play button stil don't want to be pushed ;(

ADDITION #2

Here are messages I get after I execute the code above.

Answer 1

To let Winamp play/pause the current song, simply send it a Play/Pause message, which is long 40046 as defined in winamp.h . You'll have to send it in a WM_WA_IPC message (which is a user message, also known as WM_USER ) instead of a WM_LBUTTONDOWN .

So, something like this should do:

SendMessage(parentWindow, WM_USER, IntPtr.Zero, new IntPtr(40046));

More info here , here and here .

Answer 2

well, to simulate mouse click in winamp window you need to do exactly what is happening when you just click it with hardware mouse. The following was found out from spy++:

So to emulate mouse click we need to:

1. Call twice PostMessage() instead of SendMessage() because "P" in the screenshot above tells us that the message was got from the PostMessage() (according to msdn )
2. The parameters should be the following for the first PostMessage():
   - First parameter - Winamp window handle (in our case it is 005E09FA )
   - Second parameter - the winapi code of the event, when left mouse button is down( WM_LBUTTONDOWN = 0x201 )
   - Third parameter - MK_LBUTTON = 0x1 (It seems, it is necessary to pass it. It would be great if somebody can explain why)
   - Fourth parameter - Coordinates of the point where you want to press mouse button (this parameter can be calculated as following - ((Y << 16) | X) )
3. The parameters should be the following for the second PostMessage():
   - First parameter - Winamp window handle (in our case it is 005E09FA )
   - Second parameter - the winapi code of the event, when left mouse button is up ( WM_LBUTTONUP = 0x202 )
   - Third parameter - 0 or IntPtr.Zero
   - Fourth parameter - Coordinates of the point where you want to release mouse button (need to be as in first PostMessage() if you want your click to have an effect; this parameter can be calculated as following - ((Y << 16) | X) )

Thanks everyone for their input!

Answer 3

您应该使用SendInput()将输入发送到其他应用程序，而不是伪造窗口消息。