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
最佳答案
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")
关于Python - 如何将 "an OS-level handle to an open file"转换为文件对象？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/168559/