这个是逼到没办法，C#那一套，一点基本没有。
还好，网上找到例程，可以指定帐户启动进程，但愿可以摆脱WIN SERVICE启动产生的SESSION 0 隔离问题。
因为这个问题，以SERVICE启动的进程的程序，没办法打开正常的程序窗口。
from ctypes import * from ctypes.wintypes import * INVALID_HANDLE_VALUE = -1 CREATE_UNICODE_ENVIRONMENT = 0x00000400 CData = Array.__base__ LPBYTE = POINTER(BYTE) class PROCESS_INFORMATION(Structure): '''http://msdn.microsoft.com/en-us/library/ms684873''' _fields_ = [ ('hProcess', HANDLE), ('hThread', HANDLE), ('dwProcessId', DWORD), ('dwThreadId', DWORD), ] LPPROCESS_INFORMATION = POINTER(PROCESS_INFORMATION) class STARTUPINFOW(Structure): 'http://msdn.microsoft.com/en-us/library/ms686331' _fields_ = [ ('cb', DWORD), ('lpReserved', LPWSTR), ('lpDesktop', LPWSTR), ('lpTitle', LPWSTR), ('dwX', DWORD), ('dwY', DWORD), ('dwXSize', DWORD), ('dwYSize', DWORD), ('dwXCountChars', DWORD), ('dwYCountChars', DWORD), ('dwFillAttribute', DWORD), ('dwFlags', DWORD), ('wShowWindow', WORD), ('cbReserved2', WORD), ('lpReserved2', LPBYTE), ('hStdInput', HANDLE), ('hStdOutput', HANDLE), ('hStdError', HANDLE), ] LPSTARTUPINFOW = POINTER(STARTUPINFOW) # http://msdn.microsoft.com/en-us/library/ms682431 windll.advapi32.CreateProcessWithLogonW.restype = BOOL windll.advapi32.CreateProcessWithLogonW.argtypes = [ LPCWSTR, # lpUsername LPCWSTR, # lpDomain LPCWSTR, # lpPassword DWORD, # dwLogonFlags LPCWSTR, # lpApplicationName LPWSTR, # lpCommandLine (inout) DWORD, # dwCreationFlags LPCWSTR, # lpEnvironment (force Unicode) LPCWSTR, # lpCurrentDirectory LPSTARTUPINFOW, # lpStartupInfo LPPROCESS_INFORMATION, # lpProcessInfo (out) ] def CreateProcessWithLogonW( lpUsername=None, lpDomain=None, lpPassword=None, dwLogonFlags=0, lpApplicationName=None, lpCommandLine=None, dwCreationFlags=0, lpEnvironment=None, lpCurrentDirectory=None, startupInfo=None ): if (lpCommandLine is not None and not isinstance(lpCommandLine, CData) ): lpCommandLine = create_unicode_buffer(lpCommandLine) dwCreationFlags |= CREATE_UNICODE_ENVIRONMENT if startupInfo is None: startupInfo = STARTUPINFOW(sizeof(STARTUPINFOW)) processInformation = PROCESS_INFORMATION( INVALID_HANDLE_VALUE, INVALID_HANDLE_VALUE) success = windll.advapi32.CreateProcessWithLogonW( lpUsername, lpDomain, lpPassword, dwLogonFlags, lpApplicationName, lpCommandLine, dwCreationFlags, lpEnvironment, lpCurrentDirectory, byref(startupInfo), byref(processInformation)) if not success: raise WinError() return processInformation if __name__ == '__main__': pi = CreateProcessWithLogonW( "wahaha", ".", "wahaha", 0, None, "C:\\Windows\\notepad.exe") print(pi.dwProcessId)
