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

Daryl_Spitzer_4766 0
Viewed 21678

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
fdopen mkstemp temporary-files python
6 Answers

You can use 

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

to write to the handle.

If you want to open the handle for writing you need to add the "w" mode

f = os.fdopen(tup[0], "w")
f.write("foo")
Peter_Hoffmann_7200

You forgot to specify the open mode ('w') in fdopen(). The default is 'r', causing the write() call to fail.

I think mkstemp() creates the file for reading only. Calling fdopen with 'w' probably reopens it for writing (you can reopen the file created by mkstemp).

efotinis_123200 
temp = tempfile.NamedTemporaryFile(delete=False)
temp.file.write('foo\n')
temp.close()
user_1050660

What's your goal, here? Is tempfile.TemporaryFile inappropriate for your purposes?

Alex_Coventry_19412130
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