Python - 如何将 "an OS-level handle to an open file"转换为文件对象？

Question

tempfile.mkstemp()返回:

a tuple containing an OS-level handle to an open file (as would be returned by os.open()) and the absolute pathname of that file, in that order.

如何将该操作系统级句柄转换为文件对象？

documentation for os.open()状态:

To wrap a file descriptor in a "file object", use fdopen().

所以我尝试了:

>>> import tempfile
>>> tup = tempfile.mkstemp()
>>> import os
>>> f = os.fdopen(tup[0])
>>> f.write('foo\n')
Traceback (most recent call last):
  File "<stdin>", line 1, in ?
IOError: [Errno 9] Bad file descriptor

Answer 1

你可以使用

os.write(tup[0], "foo\n")

写入句柄。

如果要打开句柄进行书写需要添加"w"模式

f = os.fdopen(tup[0], "w")
f.write("foo")