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In Python , If I am using "wget" to download a file using os.system("wget ), it shows on the screen like:

 Resolving...

 Connecting to ...

 HTTP request sent, awaiting response...

 100%[====================================================================================================================================================================>] 19,535,176  8.10M/s   in 2.3s

etc on the screen.

What can I do to save this output in some file rather than showing it on the screen ?

Currently I am running the command as follows:

theurl = "< file location >"
downloadCmd = "wget "+theurl
os.system(downloadCmd)
Shadow
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nsh
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4 Answers4

20

The os.system functions runs the command via a shell, so you can put any stdio redirects there as well. You should also use the -q flag (quiet) to wget. 

cmd = "wget -q " + theurl + " >/dev/null 2>&1"

However, there are better ways of doing this in python, such as the pycurl wrapper for libcurl, or the "stock" urllib2 module.

Keith
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4

To answer your direct question, and as others have mentioned, you should strongly consider using the subprocess module. Here's an example:

from subprocess import Popen, PIPE, STDOUT

wget = Popen(['/usr/bin/wget', theurl], stdout=PIPE, stderr=STDOUT)
stdout, nothing = wget.communicate()    

with open('wget.log', 'w') as wgetlog:
    wgetlog.write(stdout)

But, no need to call out to the system to download a file, let python do the heavy lifting for you.

Using urllib, 

try: 
    # python 2.x
    from urllib import urlretrieve
except ImportError:
    # python 3.x
    from urllib.request import urlretrieve

urlretrieve(theurl, local_filename)

Or urllib2,

import urllib2

response = urllib2.urlopen(theurl)
with open(local_filename, 'w') as dl:
    dl.write(response.read())

local_filename is the destination path of your choosing. It is sometimes possible to determine this value automatically, but the approach depends on your circumstance.

Marty
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  • I am getting the following error: File "python.py", line 129 with open('wget.log', 'w') as wgetlog: ^ SyntaxError: invalid syntax – nsh Aug 20 '11 at 17:36
  • Without using the keyword "with" and "as", it works successfully .... wget = Popen(['/usr/bin/wget', theurl], stdout=PIPE, stderr=STDOUT) stdout, nothing = wget.communicate() wgetlog = open('wget.log', 'w') wgetlog.write(stdout) – nsh Aug 20 '11 at 17:42
  • @nsh Yes, sorry. The [with statement](http://docs.python.org/reference/compound_stmts.html#the-with-statement) is enabled by default in python 2.6 and higher, you can enable it in python 2.5 by adding `from __future__ import with_statement` as the first import. If you're using python 2.4 or previous see http://stackoverflow.com/questions/3770348/how-to-safely-open-close-files-in-python-2-4 – Marty Aug 20 '11 at 20:38
1

As others have noted, you can use Python native library modules to do your I/O, or you can modify the command line to redirect the output.

But for full control over the output, the best thing is to use the Python subprocess module instead of os.system(). Using subprocess would let you capture the output and inspect it, or feed arbitrary data into standard input.

When you want a quick-and-dirty way to run something, use os.system(). When you want full control over how you run something, use subprocess.

steveha
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0

The wget process is just writing to STDOUT (and perhaps STDERR if something bad happens) and these are still "wired" to the terminal.

To get it to stop doing this, redirect (or close) said filehandles. Look at the subprocess module which allows configuring said filehandles when starting a process. (os.system just leaves the STDOUT/STDERR of the spawned process alone and thus they are inherited but the subprocess module is more flexible.)

See Working with Python subprocess - Shells, Processes, Streams, Pipes, Redirects and More for lots of nice examples and explanations (it introduces the concepts of STDIN/STDOUT/STDERR and works from there).

There are likely better ways to handle this than using wget -- but I'll leave such to other answers.

Happy coding.

  • I imported the module subprocess. Then I am using it as follows ----->process = subprocess.Popen(downloadCmd, shell=False, stdout=subprocess.PIPE) But it is giving me the error:File "/usr/lib64/python2.4/subprocess.py", line 550, in __init__ errread, errwrite) File "/usr/lib64/python2.4/subprocess.py", line 996, in _execute_child raise child_exception OSError: [Errno 2] No such file or directory – nsh Aug 20 '11 at 02:52