Develop Reference

Running Batch File from Python

I am trying to run a batch file through python; however, it is not recognizing the path. It stops reading the path after the space between 'Practice' and 'Folder'. How can I fix this? I've tried the r and using forward and backward slashes. Any help would be awesome. Thank you!
import os
Practice = r"C:\Users\Username\Desktop\Practice Folder\Practice.bat"
os.system(Practice)
'C:\Users\Username\Desktop\Practice' is not recognized as an internal
or external command, operable program or batch file.

Change working directory to the script directory as you are using some relative redirection paths. Pushd changes current directory to any drive and can map network drives. The && chains commands and only runs the right hand command if the left hand command succeeds. %UserProfile% is a standard environmental variable which is usually better then using a fixed path of C:\Users\Username.
import os
Practice = r'pushd "%UserProfile%\Desktop\Practice Folder" && Practice.bat'
os.system(Practice)

Try using call from subprocess module.
You need to enclose the command only in double quotes.
from subprocess import call
call(r'"C:\Users\Username\Desktop\Practice Folder\Practice.bat"')
(Notice the order of placing quotes...)
This would even work with os.system() provided you take care the order of quotation marks.
from os import system
system(r'"C:\Users\Username\Desktop\Practice Folder\Practice.bat"')
This should help fix your problem.

You probably need to use two types of quotation marks e.g.
import os
Practice = r"'C:\Users\Username\Desktop\Practice Folder\Practice.bat'"
os.system(Practice)
As it is, your string does not contain quotation marks - you need to include quotation marks within your string or else Windows will think that Folder\Practice.bat is an argument to the command rather than a continuation of the file path

Try this
import os
Practice = os.path.abspath(r"C:\Users\Username\Desktop\Practice Folder\Practice.bat")
Edit:
Something like this worked for me
os.system(r'"C:\Users\Username\Desktop\Practice Folder\Practice.bat"')

How to escape a spacebar in a path name with subprocess?

I'm trying to convert a file from .m4a to .mp3 using ffmpeg and I need to access to the music folder.
The path name of this folder is : C:\\Users\A B\Desktop\Music
I can't access it with subprocess.call() because only C:\\Users\A gets recognized. The white space is not processed.
Here's my python script :
import constants
import os
import subprocess
path = 'C:\\Users\A B\Desktop\Music'
def main():
files = sorted(os.listdir(path), key=lambda x: os.path.getctime(os.path.join(path, x)))
if "Thumbs.db" in files: files.remove("Thumbs.db")
for f in files:
if f.lower()[-3:] == "m4a":
process(f)
def process(f):
inFile = f
outFile = f[:-3] + "mp3"
subprocess.call('ffmpeg -i {} {} {}'.format('C:\\Users\A B\Desktop\Music', inFile, outFile))
main()
When I run it I get an error that states :
C:\Users\A: No such file or directory
I wonder if someones knows how to put my full path name (C:\Users\A B\Desktop\Music) in subprocess.call() ?

Beforehand edit: spaces or not, the following command line -i <directory> <infilename> <outfilename> is not correct for ffmpeg since it expects the -i option, then input file and output file, not a directory first. So you have more than one problem here (which explains the "permission denied" message you had, because ffmpeg was trying to open a directory as a file!)
I suppose that you want to:
read all files from directory
convert them all to a file located in the same directory
In that case, you could add quotes to your both input & output absolute files like this:
subprocess.call('ffmpeg -i "{0}\{1}" "{0}\{2}"'.format('C:\\Users\A B\Desktop\Music', inFile, outFile))
That would work, but that's not the best thing to do: not very performant, using format when you already have all the arguments already, you may not have knowledge of other characters to escape, etc... don't reinvent the wheel.
The best way to do it is to pass the arguments in a list so subprocess module handles the quoting/escaping when necessary:
path = r'C:\Users\A B\Desktop\Music' # use raw prefix to avoid backslash escaping
subprocess.call(['ffmpeg','-i',os.path.join(path,inFile), os.path.join(path,outFile)])
Aside: if you're the user in question, it's even better to do:
path = os.getenv("USERPROFILE"),'Desktop','Music'
and you could even run the process in the path directory with cwd option:
subprocess.call(['ffmpeg','-i',inFile, outFile],cwd=path)
and if you're not, be sure to run the script with elevated privileges or you won't get access to another user directory (read-protected)

subprocess.call() to remove files

I want to remove all the files and directories except for some of them by using
`subprocess.call(['rm','-r','!(new_models|creat_model.py|my_mos.tit)'])`
but it gives back information
rm: cannot remove `!(new_models|creat_model.py|my_mos.tit)': No such file or directory
how can I fix this? Thanks

If you use that rm command on the command line the !(…|…|…) pattern is expanded by the shell into all file names except those in the pattern before calling rm. Your code calls rm directly so rm gets the shell pattern as a file name and tries to delete a file with that name.
You have to add shell=True to the argument list of subprocess.call() or actually code this in Python instead of calling external commands. Downside: That would be more than one line. Upside: it can be done independently from external shells and system dependent external programs.

An alternative to shell=True could be the usage of glob and manual filtering:
import glob
files = [i for i in glob.glob("*") if i not in ('new_models', 'creat_model.py', 'my_mos.tit')]
subprocess.call(['rm','-r'] + files)
Edit 4 years later:
Without glob (of which I don't remember why I suggested it):
import os
files = [i for i in os.listdir() if i not in ('new_models', 'creat_model.py', 'my_mos.tit')]
subprocess.call(['rm','-r'] + files)

Code to remove all png
args = ('rm', '-rf', '/path/to/files/temp/*.png')
subprocess.call('%s %s %s' % args, shell=True)

Trouble extracting zip in python over ftp

I'm trying to unzip a file from an FTP site. I've tried it using 7z in a subprocess as well as using 7z in the older os.system format. I get closest however when I'm using the zipfile module in python so I've decided to stick with that. No matter how I edit this I seem to get one of two errors so here are both of them so y'all can see where I'm banging my head against the wall:
z = zipfile.ZipFile(r"\\svr-dc\ftp site\%s\daily\data1.zip" % item)
z.extractall()
NotImplementedError: compression type 6 (implode)
(I think this one is totally wrong, but figured I'd include.)
I seem to get the closest with the following:
z = zipfile.ZipFile(r"\\svr-dc\ftp site\%s\daily\data1.zip" % item)
z.extractall(r"\\svr-dc\ftp site\%s\daily\data1.zip" % item)
IOError: [Errno 2] No such file or directory: '\\\\svr-dc...'
The catch with this is that it is actually giving me the first file name in the zip. I can see the file AJ07242013.PRN at the end of the error so I feel closer because it's at least getting to the point of reading the contents of the zip file.
Pretty much any iteration of this that I try gets me one of those two errors, or a syntax error but that's easily addressed and not my primary concern.
Sorry for being so long winded. I'd love to get this working, so let me know what you think I need to do.
EDIT:
So 7z has finally been added to the path and is running through without any errors with both the subprocess as well as os.system. However, I still can't seem to get anything to unpack. It looks to me, from all I've read in the python documentation that I should be using the subprocess.communicate() module to extract this file but it just won't unpack. When I use os.system it keeps telling me that it cannot find the archive.
import subprocess
cmd = ['7z', 'e']
sp = subprocess.Popen(cmd, stderr=subprocess.STDOUT, stdout=subprocess.PIPE)
sp.communicate('r"\C:\Users\boster\Desktop\Data1.zip"')
I don't think that sp.communicate is right but if I add anything else to it I have too many arguments.

python's zipfile doesn't support compression type 6 (imploded) so its simply not going to work. In the first case, that's obvious from the error. In the second case, things are worse. The parameter for extractfile is an alternate unzip directory. Since you gave it the name of your zip file, a directory of the same name can't be found and zipfile gives up before getting to the not-supported problem.
Make sure you can do this with 7z on the command line, try implementing subprocess again and ask for help on that technique if you need it.
Here's a script that will look for 7z in the usual places:
import os
import sys
import subprocess
from glob import glob
print 'python version:', sys.version
subprocess.call('ver', shell=True)
print
if os.path.exists(r'C:\Program Files\7-Zip'):
print 'have standard 7z install'
if '7-zip' in os.environ['PATH'].lower():
print '...and its in the path'
else:
print '...but its not in the path'
print
print 'find in path...'
found = 0
for p in os.environ['PATH'].split(os.path.pathsep):
candidate = os.path.join(p, '7z.*')
for fn in glob(candidate):
print ' found', fn
found += 1
print
if found:
print '7z located, attempt run'
subprocess.call(['7z'])
else:
print '7z not found'

Accoring to the ZipFile documentation, you might be better off copying the zip first to your working directory. (http://docs.python.org/2/library/zipfile#zipfile.ZipFile.extract)
If you have problems copying, you might want to store the zip in a path with no spaces or protect your code against spaces by using os.path.
I made a small test in which I used os.path.abspath to make sure I had the proper path to my zip and it worked properly.
Also make sure that for extractall the path that you specify is the path where the zip content will be extracted. (If a folder that is specified is not created, it will be created automatically) Your files will be extracted in your current working directory (CWD) if no parameter is passed to extractall.
Cheers!

Managed to get this to work without using the PIPE functionality as subprocess.communicate wouldn't unpack the files. Here was the solution using subprocess.call. Hope this can help someone in the future.
def extract_data_one():
for item in sites:
os.chdir(r"\\svr-dc\ftp site\%s\Daily" % item)
subprocess.call(['7z', 'e', 'data1.zip', '*.*'])

Star in sys.argv in python

I am attempting to write a script that utilises sys.argv to wrap the scp command. The idea, is that you will be able to run: pyscp folder/* host but if I run this script with those arguments:
import sys
for arg in sys.argv:
print arg
I get a list of all the folders inside folder:
pyscp.py
folder/0
folder/1
folder/2
folder/3
folder/4
folder/5
folder/67
folder/8
folder/9
host

Assuming a UNIXoid operating system: The shell is expanding the * into the matching files. Try to call your script like
pyscp "folder/*" host
The quotes keep the shell from interpreting the * character.

If you do not escape the asterisk, the shell is performing filename expansion for you. The pattern including the asterisk becomes replaced with an alphabetically sorted list of file names matching the pattern before your Python program becomes executed. You can prevent the shell from performing filename expansion using e.g. single quotes, i.e.
pyscp 'folder/*' hostname
You can then do this yourself within Python using the glob module and control things the way you want it.

The shell is expanding the file list for you. You can leverage this by allowing multiple parameters in the command.
import sys
files = sys.argv[1:-1]
host = sys.argv[-1]
Now you have a more flexible program that lets caller jump through whatever hoops he wants for the transfer, like maybe all text files in folder1 plus anything that's changed in the last day in folder2 (on a linux machine):
pyscp folder1/*.txt `find -mtime -1` example.com