Python scp copy file with spaces in filename

Question

I'm trying to copy files in local network with scp. It's working well with filenames without spaces, but it crash with. I've tried to replace " " with "\ " as this exemple, but it don't work. Here is my code:

def connection(locals):
         a = (int(re.search(br'(\d+)%$', locals['child'].after).group(1)))
         print a
         perc = (Decimal(a)/100)
         print (type(perc)), perc
         while gtk.events_pending():
             gtk.main_iteration()
         FileCopy.pbar.set_text("Copy of the file in the Pi...   " + str(a) + "%")
         while gtk.events_pending():
             gtk.main_iteration()
         FileCopy.pbar.set_fraction(perc)

file_pc = "/home/guillaume/folder/a very large name of file with space .smthg"
file_pi = "pi@192.168.X.X:/home/pi/folder/a very large name of file with space .smthg"

if " " in file_pc:
   file_pc = fichier_pc.replace(" ", '\\\ ')   # tried '\\ ' or '\ '
   file_pi = fichier_pi.replace(" ", '\\\ ')   # but no way
else:
   pass
command = "scp %s %s" % tuple(map(pipes.quote, [file_pc, file_pi]))
pexpect.run(command, events={r'\d+%': connection}) # this command is using to get the %

How can I fix this problem ? Thanks

related: [How to escape spaces in path during scp copy in linux?](http://stackoverflow.com/q/19858176/4279) — jfs, Mar 29 '14 at 13:12

score 3 · Answer 1 · answered Mar 29 '14 at 09:58

3

Use subprocess module and/or shlex.split():

import subprocess
subprocess.call(['scp', file_pc, file_pi])

and you don't need to worry about escaping or quoting anything

answered Mar 29 '14 at 09:58

m.wasowski

6,329
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No, I can't. I want to get the output of the command, to display the progressbar. (see [here](http://stackoverflow.com/questions/22687296/python-copy-file-in-local-network-linux-linux-and-output) – Guillaume Mar 29 '14 at 10:06
output is displayed on stdout. If you wan't it captured to react in realtime... well, it can be done, but it's a lot of hassle. You need threads, reopening fd descriptor for process stdout with proper flags etc. It would be better to check `paramiko` module, but that's completely different question. – m.wasowski Mar 29 '14 at 10:19
I tried paramiko, but, there is a bug with big files (see [here](http://stackoverflow.com/questions/22708942/python-paramiko-module-error-with-callback)). As I say, the script works fine with no spaces in the name. – Guillaume Mar 29 '14 at 12:59
@m.wasowski: read [my answer to the question that Guillaume've linked](http://stackoverflow.com/a/22695208/4279). It shows that `scp` prints *nothing* if stdout is a pipe. Though it works with pseudo-tty provided by `pexpect` module (there are no buffering issues in this case). The issue is how to pass filenames with spaces to scp command – jfs Mar 29 '14 at 13:14
Ok, I give another solution when i get home – m.wasowski Mar 29 '14 at 13:32

score 2 · Accepted Answer · answered Mar 29 '14 at 13:19

2

You may keep local file file_pc as is (pipes.quote will escape the spaces). The remote file should be changed:

import pipes

file_pi = 'pi@192.168.X.X:/home/pi/folder/file with space.smth'
host, colon, path = file_pi.partition(':')
assert colon
file_pi = host + colon + pipes.quote(path)

i.e., user@host:/path/with space should be changed to user@host:'/path/with space'

answered Mar 29 '14 at 13:19

jfs

399,953
195
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1,670

You're strong. Thank you! – Guillaume Mar 29 '14 at 13:47

score 0 · Answer 3 · answered Mar 29 '14 at 13:26

0

You might want to look into fabric, a Python library that streamlines the use of SSH.

from fabric.state import env
from fabric.operations import get

env.user = 'username'
env.key_filename = '/path/to/ssh-key'

get('/remote_path/*', 'local_path/')

answered Mar 29 '14 at 13:26

rubayeet

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`fabric` is implemented on top of [`paramiko` that breaks for large files](http://stackoverflow.com/questions/22708942/python-paramiko-module-error-with-callback) – jfs Mar 29 '14 at 13:49

Python scp copy file with spaces in filename

3 Answers3

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