如何在python中获取打开文件的win32句柄? [英] How to get a win32 handle of an open file in python?
问题描述
我确定这在某处有记录,但我找不到它...
我的代码正在从另一个库中获取一个 python 对象(我无法修改),我需要在它上面调用一些 win32 api 函数.
Python 从 file.fileno() 返回一些不是操作系统级句柄的东西,我猜它给出了 MSVCRT 的文件号.
<预><代码>>>>ctypes.windll.kernel32.CreateFileA('test',0x80000000L,1,None,3,0,0)1948 # <- 手柄>>>file('test','r').fileno()4 # <- 不是句柄如何将其转换为真正的 win32 句柄?
我找到了答案:
<预><代码>>>>msvcrt.get_osfhandle(a.fileno())1956 # 有效的句柄这实际上记录在 http://docs.python.org/library/msvcrt.html ,不知道我是怎么错过的.
I'm sure this is documented somewhere but i can't find it...
My code is getting a python object from another library (that i can't modify), and i need to call some win32 api functions on it.
Python returns something that isn't the os-level handle from file.fileno(), my guess is that it gives MSVCRT's fileno.
>>> ctypes.windll.kernel32.CreateFileA('test',0x80000000L,1,None,3,0,0)
1948 # <- HANDLE
>>> file('test','r').fileno()
4 # <- not a HANDLE
How do i convert it into a real win32 handle?
I found the answer:
>>> msvcrt.get_osfhandle(a.fileno())
1956 # valid HANDLE
This is actually documented on http://docs.python.org/library/msvcrt.html , no idea how i missed it.
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