如何获取当前活动应用程序窗口的路径？

Question

I want to get the path to the executable file of the current active Window.

I tried:

var
  WindowModuleFileName : array[0..100] of Char;  
  sourceWindow: Hwnd;      
begin
  sourceWindow := GetActiveWindow;
  GetWindowModuleFileName(sourceWindow, PChar(WindowModuleFileName), sizeof(WindowModuleFileName));    
  ShowMessage(WindowModuleFileName);    
end;

But it returned correct answer only when my application window was active. What am I doing wrong?

Answer 1

You can't use GetWindowModuleFileName to locate files for other processes than your own, as stated on GetModuleFileName MSDN :

Retrieves the fully-qualified path for the file that contains the specified module. The module must have been loaded by the current process.

To locate the file for a module that was loaded by another process, use the GetModuleFileNameEx function.

Therefore, you have to use GetModuleFileNameEx combined with GetWindowThreadProcessId / GetForegroundWindow . This will return you what you need:

uses
  Winapi.Windows, Winapi.PsAPI, System.SysUtils;

function GetCurrentActiveProcessPath: String;
var
  pid     : DWORD;
  hProcess: THandle;
  path    : array[0..4095] of Char;
begin
  GetWindowThreadProcessId(GetForegroundWindow, pid);

  hProcess := OpenProcess(PROCESS_QUERY_INFORMATION or PROCESS_VM_READ, FALSE, pid);
  if hProcess <> 0 then
    try
      if GetModuleFileNameEx(hProcess, 0, @path[0], Length(path)) = 0 then
        RaiseLastOSError;

      result := path;
    finally
      CloseHandle(hProcess);
    end
  else
    RaiseLastOSError;
end;

Answer 2

GetActiveWindow gets the window handle to the active window that belongs to the calling thread only.

You need to use GetForegroundWindow() function instead of GetActiveWindow() .