Wildcard not working in subprocess call using shlex

Question

Language: Python v2.6.2 OS: AIX 5.3 I'm using Python to restore some files from a backup to a test system - all commands are called in the manner below, however some just plain don't want to work. ``` #!/usr/bin/python import subprocess, shlex cmd = 'sudo rm -rf /work/TEST/*' arg = shlex.split(cmd) # This does not work p = subprocess.Popen(arg) # This, however, works just fine p = subprocess.Popen(cmd, shell=True) ``` If I remove the *'s from the commands they work fine (well, they work as they should without the wildcards, which is unfortauntely not what I want). I really do not want to use shell=True for obvious security reasons, however there are a couple of other commands that basically do the same thing. If there is a wildcard in the command it just won't work - it executes without error, just doesn't do anything. Interestingly the following command (parsed through shlex): ``` sudo mv /work/testrestore/production/* /work/TESTC ``` Produces the following: mv: 0653-401 Cannot renam…

Accepted Answer

For replacing the `*` with what it means, you either need the shell or you need the `glob` module. So the easiest way would be `shell=True` (if the command is constant, I do not see any security holes).

Another approach would be

```
#!/usr/bin/python
import subprocess
import shlex
import glob

cmd = 'sudo rm -rf /work/TEST/*'
arg = shlex.split(cmd)
arg = arg[:-1] + glob.glob(arg[-1])

# This should work now
p = subprocess.Popen(arg)
```

or, if you would nevertheless append the path by yourself,

```
cmd = 'sudo rm -rf'
basearg = shlex.split(cmd)
arg = basearg + glob.glob(path+"/*")
```

Wildcard not working in subprocess call using shlex

Solution

Problem