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Groups > comp.lang.python > #100792
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| From | eryk sun <eryksun@gmail.com> |
| Newsgroups | comp.lang.python |
| Subject | Re: What interface is a ‘Popen.stdout’ presenting? |
| Date | Wed, 23 Dec 2015 22:30:02 -0600 |
| Lines | 19 |
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| References | <85d1twzlj2.fsf@benfinney.id.au> <858u4kzkyp.fsf@benfinney.id.au> |
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On Wed, Dec 23, 2015 at 7:36 PM, Ben Finney <ben+python@benfinney.id.au> wrote:
> So how do I get from a Python 2 ‘file’ object, to whatever
> ‘io.TextIOWrapper’ wants?
I would dup the file descriptor and close the original file. Then open
the file descriptor using io.open:
>>> p = subprocess.Popen(['uname'], stdout=subprocess.PIPE)
>>> fd = os.dup(p.stdout.fileno())
>>> fd
4
>>> p.stdout.close()
>>> fout = io.open(fd, 'r')
>>> fout
<_io.TextIOWrapper name=4 encoding='UTF-8'>
>>> fout.read()
u'Linux\n'
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Re: What interface is a ‘Popen.stdout’ presenting? eryk sun <eryksun@gmail.com> - 2015-12-23 22:30 -0600
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