Send key to JUMP TO FILE winamp window

Catalin

Hi,
I'm trying to send key stroke to winamp Jump to File window, using c#, when is not in foreground but I don't know how.
So far I managed to send keys to winamp like "c" key to toggle play/pause using this example (this example work even when winamp is minimized):

using System;
using System.Diagnostics;
using System.Runtime.InteropServices;

namespace WindowsFormsApplication1
{
    class Test
    {

        [DllImport("User32.Dll", EntryPoint = "PostMessageA")]
        static extern bool PostMessage(IntPtr hWnd, uint msg, int wParam, int lParam);

        [DllImport("user32.dll", CharSet = CharSet.Auto)]
        public static extern IntPtr FindWindow([MarshalAs(UnmanagedType.LPTStr)] string lpClassName,
                                                [MarshalAs(UnmanagedType.LPTStr)] string lpWindowName);

       public static void com()
        {
            IntPtr hwnd = FindWindow("Winamp v1.x", null);
            PostMessage(hwnd, 0x100, 0x43, 0);
        }
    }
}

0x43 is the Virtual-Key Code for „C" key.

Can someone help me with this problem and give me an example of haw I can send keys to Jump to File window in winamp?

6 Replies

Know the answer? Post it — somebody with the same question will find it here.

Vulpes14y ago
Accepted answer
I think the problem here is that "Edit" is not a child of Winamp's main window but a child of some other window (apparently one with the class name "#32770") down the chain.

What you'll therefore need to do is to repeatedly call FindWindowEx for each level of child window until you eventually drill down to "Edit".
Catalin14y ago
Hello again.
Using your advice I managed to send keys to "Jump to File" window.
Below is the script that works perfect for me:
using System;
using System.Diagnostics;
using System.Runtime.InteropServices;
namespace WindowsFormsApplication1
{
    class test
    {
        [DllImport("User32.Dll", EntryPoint = "PostMessageA")]
        static extern bool PostMessage(IntPtr hWnd, uint msg, int wParam, int lParam);
        [DllImport("user32.dll", CharSet = CharSet.Auto)]
        public static extern IntPtr FindWindow([MarshalAs(UnmanagedType.LPTStr)] string lpClassName,
                                               [MarshalAs(UnmanagedType.LPTStr)] string lpWindowName);
        [DllImport("user32.dll", CharSet = CharSet.Auto)]
        static extern IntPtr FindWindowEx(IntPtr hwndParent, IntPtr hwndChildAfter, string lpszClass, string lpszWindow);
        public static void comm()
        {
            IntPtr hwnd = FindWindow("BaseWindow_RootWnd", null);
            IntPtr chwnd1 = FindWindowEx(hwnd, IntPtr.Zero, "BaseWindow_RootWnd", null);
            IntPtr chwnd2 = FindWindowEx(chwnd1, IntPtr.Zero, "Winamp Gen", null);
            IntPtr chwnd3 = FindWindowEx(chwnd2, IntPtr.Zero, "#32770", null);
            IntPtr chwnd4 = FindWindowEx(chwnd3, IntPtr.Zero, "Edit", null);
            PostMessage(chwnd4, 0x100, 0x43 , 0);


        }

    }
}
0x43 is the Virtual-Key Code for „C" key.

Thank you very much for your answer.
Catalin14y ago
I obtained the name of the class using Winspector.
This is what I found with Winspector:
- 000B099A: BaseWindow_RootWnd „Jump to file"
        - 00090820: BaseWindow_RootWnd „Jump to file"
                - 004E0242: Winamp Gen „Jump to file"
                        - 001707CC: #32770 „Jump to file"
                                - 00200806: Edit
I changed my script into this but with no result.
using System;
using System.Diagnostics;
using System.Runtime.InteropServices;
namespace WindowsFormsApplication1
{
    class Test2
    {

        [DllImport("User32.Dll", EntryPoint = "PostMessageA")]
        static extern bool PostMessage(IntPtr hWnd, uint msg, int wParam, int lParam);
        [DllImport("user32.dll", CharSet = CharSet.Auto)]
        public static extern IntPtr FindWindow([MarshalAs(UnmanagedType.LPTStr)] string lpClassName,
                                               [MarshalAs(UnmanagedType.LPTStr)] string lpWindowName);
        [DllImport("user32.dll", CharSet = CharSet.Auto)]
        static extern IntPtr FindWindowEx(IntPtr hwndParent, IntPtr hwndChildAfter, string lpszClass, string lpszWindow);

       public static void comm()
        {
            IntPtr hwnd = FindWindow("Winamp v1.x", null);
            IntPtr chwnd = FindWindowEx(hwnd, IntPtr.Zero, "Edit", null);
            PostMessage(chwnd, 0x100, 0x43, 0);
        }
    }
}

Maybe these are banal questions but I'm very new in c#
Thank you very much for your time.
Vulpes14y ago
Well, you'll need to use FindWindowEx to get the handle for the child window and then send it the keys using PostMessage.

I imagine you've obtained the window class name using Spy++.

The code will be something like this:

[DllImport("user32.dll", CharSet = CharSet.Auto)]
static extern IntPtr FindWindowEx(IntPtr hwndParent, IntPtr hwndChildAfter, string lpszClass, string lpszWindow);

// ...

IntPtr chwnd = FindWindowEx(hwnd, null, "Edit", null);
Catalin14y ago
Thanks for the answer.

What I want is to send keys to the window (JUMP TO FILE) not to open it. I'm trying to make an application that search a melody in winamp play list by sending keys to this window.
From what I know I have to send keys to a child class of wimamp which is called "Edit", but I don't know how.

I found some examples but it didn't work for me.
I want to use "PostMessage" because I want to be able to sent keys even the window is not in focus.

Thanks.
Vulpes14y ago
According to this link:

http://blog.winamp.com/2008/08/14/keyboard-shortcuts/

the keyboard shortcut for Jump To File is 'J'.

So, I'd send the virtual keycode for 'J' which is 0x4A.