如何获取开始菜单程序目录的路径？

Question

…对于当前用户？对所有用户？在

我在做一个小程序，需要在开始菜单中创建链接。目前我的硬编码如下，但它只适用于英语地区，例如，它应该是“Startmenü”在德语。什么是更清洁、更便携的方法？在

OUR_STARTMENU = os.environ['ALLUSERSPROFILE'] + '\Start Menu\Programs\Our Stuff'

谢谢你

Answer 1

一个朋友，卢克·平纳Environment.gov.au，通过电子邮件给出了一个使用核心模块（python2.5+）的解决方案。它被认为是多语种的，因为API调用返回的是unicode。在Win7上测试了日语语言环境，并在另一台美英机器上手动更改开始菜单指向%USERPROFILE%\Startmenü

''' Get windows special folders without pythonwin
    Example:
            import specialfolders
            start_programs = specialfolders.get(specialfolders.PROGRAMS)

Code is public domain, do with it what you will. 

Luke Pinner - Environment.gov.au, 2010 February 10
'''

#Imports use _syntax to mask them from autocomplete IDE's
import ctypes as _ctypes
from ctypes.wintypes import HWND as _HWND, HANDLE as _HANDLE,DWORD as _DWORD,LPCWSTR as _LPCWSTR,MAX_PATH as _MAX_PATH, create_unicode_buffer as _cub
_SHGetFolderPath = _ctypes.windll.shell32.SHGetFolderPathW

#public special folder constants
DESKTOP=                             0
PROGRAMS=                            2
MYDOCUMENTS=                         5
FAVORITES=                           6
STARTUP=                             7
RECENT=                              8
SENDTO=                              9
STARTMENU=                          11
MYMUSIC=                            13
MYVIDEOS=                           14
NETHOOD=                            19
FONTS=                              20
TEMPLATES=                          21
ALLUSERSSTARTMENU=                  22
ALLUSERSPROGRAMS=                   23
ALLUSERSSTARTUP=                    24
ALLUSERSDESKTOP=                    25
APPLICATIONDATA=                    26
PRINTHOOD=                          27
LOCALSETTINGSAPPLICATIONDATA=       28
ALLUSERSFAVORITES=                  31
LOCALSETTINGSTEMPORARYINTERNETFILES=32
COOKIES=                            33
LOCALSETTINGSHISTORY=               34
ALLUSERSAPPLICATIONDATA=            35

def get(intFolder):
    _SHGetFolderPath.argtypes = [_HWND, _ctypes.c_int, _HANDLE, _DWORD, _LPCWSTR]
    auPathBuffer = _cub(_MAX_PATH)
    exit_code=_SHGetFolderPath(0, intFolder, 0, 0, auPathBuffer)
    return auPathBuffer.value

Answer 2

我听说过两种方法。第一个：

from win32com.shell import shell
shell.SHGetSpecialFolderPath(0,shellcon.CSIDL_COMMON_STARTMENU)

第二，使用WScript.Shell对象（源：http://www.mail-archive.com/python-win32@python.org/msg00992.html）：

^{pr2}$

另一个来源：http://blogs.msdn.com/saveenr/archive/2005/12/28/creating-a-start-menu-shortcut-with-powershell-and-python.aspx

Answer 3

此外，CSIDL_COMMON洇STARTMENU用于所有用户启动，CSIDL\u STARTMENU用于当前用户启动。在