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Python - How do I convert "an OS-level handle to an open file" to a file object?

Asked 2008-10-03T19:41:04.963
61

tempfile.mkstemp() returns:

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.

How do I convert that OS-level handle to a file object?

The documentation for os.open() states:

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

So I tried:

>>> 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
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I can't comment on the answers, so I will post my comment here:

To create a temporary file for write access you can use tempfile.mkstemp and specify "w" as the last parameter, like:

f = tempfile.mkstemp("", "", "", "w") # first three params are 'suffix, 'prefix', 'dir'...
os.write(f[0], "write something")
answered 2013-06-05T14:15:44.287

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