在执行Shell脚本时，通过python执行带有参数的Shell脚本始终采用默认值

Question

My python is supposed to execute the /home/file.sh with five parameters as shown below. However, everytime I execute my python script it ignores the parameters and the shell script executes with default values. I have also tried turning the shell to False but no success. Any ideas ? Thank you in advance!

var1='-t'
var2='m'
var3='20130105'
var4='-f'
var5='test.txt'

process=subprocess.Popen(['/home/file.sh', str(var1), str(var2), str(var3), str(var4), str(var5)],shell=True, stdout=subprocess.PIPE, stderr=subprocess.PIPE)
process.wait()
lists=process.communicate()[0]

I have tried the following but it again use the default parameters

process=subprocess.Popen(['/home/file.sh  -t m 20140105 -f test.txt')],shell=True, stdout=subprocess.PIPE, stderr=subprocess.PIPE)

Answer 1

Try like so:

process=subprocess.Popen('/home/file.sh -t m 20140105 -f test.txt', shell=True, ...)

without worrying about passing a list for your command.

Answer 2

Either you use an argument list and set shell=False , eg

process=subprocess.Popen(["./test.sh","a","b","c"], shell=False, stdout=subprocess.PIPE, stderr=subprocess.PIPE)
print process.communicate()[0] 

or you can use a string argument and let the shell parse it by setting shell=True , eg

process=subprocess.Popen("./test.sh a b c", shell=True, stdout=subprocess.PIPE, stderr=subprocess.PIPE)
print process.communicate()[0]

but you can not send a list when using shell=True .

Both will output

0: ./test.sh
1: a
2: b
3: c

if test.sh is an executable file with the following contents:

#!/bin/sh

echo 0: $0
echo 1: $1
echo 2: $2
echo 3: $3

Ususally, I prefer using shell=False to avoid having to escape the argument strings (it's trickier than it should be to get this right). Using the shell has some advantages though, such as expansion of paths, being able to use shell internals and so on.