Running bash script from within python
Question:
I have a problem with the following code:
callBash.py:
import subprocess
print "start"
subprocess.call("sleep.sh")
print "end"
sleep.sh:
sleep 10
I want the “end” to be printed after 10s. (I know that this is a dumb example, I could simply sleep within python, but this simple sleep.sh file was just as a test)
Answers:
Make sure that sleep.sh has execution permissions, and run it with shell=True:
#!/usr/bin/python
import subprocess
print "start"
subprocess.call("./sleep.sh", shell=True)
print "end"
Actually, you just have to add the shell=True argument:
subprocess.call("sleep.sh", shell=True)
But beware –
Warning Invoking the system shell with shell=True can be a security hazard if combined with untrusted input. See the warning under Frequently Used Arguments for details.
Making sleep.sh executable and adding shell=True to the parameter list (as suggested in previous answers) works ok. Depending on the search path, you may also need to add ./ or some other appropriate path. (Ie, change "sleep.sh" to "./sleep.sh".)
The shell=True parameter is not needed (under a Posix system like Linux) if the first line of the bash script is a path to a shell; for example, #!/bin/bash.
If sleep.sh has the shebang #!/bin/sh and it has appropriate file permissions — run chmod u+rx sleep.sh to make sure and it is in $PATH then your code should work as is:
import subprocess
rc = subprocess.call("sleep.sh")
If the script is not in the PATH then specify the full path to it e.g., if it is in the current working directory:
from subprocess import call
rc = call("./sleep.sh")
If the script has no shebang then you need to specify shell=True:
rc = call("./sleep.sh", shell=True)
If the script has no executable permissions and you can’t change it e.g., by running os.chmod('sleep.sh', 0o755) then you could read the script as a text file and pass the string to subprocess module instead:
with open('sleep.sh', 'rb') as file:
script = file.read()
rc = call(script, shell=True)
Adding an answer because I was directed here after asking how to run a bash script from python. You receive an error OSError: [Errno 2] file not found if your script takes in parameters. Lets say for instance your script took in a sleep time parameter: subprocess.call("sleep.sh 10") will not work, you must pass it as an array: subprocess.call(["sleep.sh", 10])
If someone looking for calling a script with arguments
import subprocess
val = subprocess.check_call("./script.sh '%s'" % arg, shell=True)
Remember to convert the args to string before passing, using str(arg).
This can be used to pass as many arguments as desired:
subprocess.check_call("./script.ksh %s %s %s" % (arg1, str(arg2), arg3), shell=True)
If chmod is not working then you can also try:
import os
os.system('sh script.sh')
# you can also use bash instead of sh
I have a problem with the following code:
callBash.py:
import subprocess
print "start"
subprocess.call("sleep.sh")
print "end"
sleep.sh:
sleep 10
I want the “end” to be printed after 10s. (I know that this is a dumb example, I could simply sleep within python, but this simple sleep.sh file was just as a test)
Make sure that sleep.sh has execution permissions, and run it with shell=True:
#!/usr/bin/python
import subprocess
print "start"
subprocess.call("./sleep.sh", shell=True)
print "end"
Actually, you just have to add the shell=True argument:
subprocess.call("sleep.sh", shell=True)
But beware –
Warning Invoking the system shell with shell=True can be a security hazard if combined with untrusted input. See the warning under Frequently Used Arguments for details.
Making sleep.sh executable and adding shell=True to the parameter list (as suggested in previous answers) works ok. Depending on the search path, you may also need to add ./ or some other appropriate path. (Ie, change "sleep.sh" to "./sleep.sh".)
The shell=True parameter is not needed (under a Posix system like Linux) if the first line of the bash script is a path to a shell; for example, #!/bin/bash.
If sleep.sh has the shebang #!/bin/sh and it has appropriate file permissions — run chmod u+rx sleep.sh to make sure and it is in $PATH then your code should work as is:
import subprocess
rc = subprocess.call("sleep.sh")
If the script is not in the PATH then specify the full path to it e.g., if it is in the current working directory:
from subprocess import call
rc = call("./sleep.sh")
If the script has no shebang then you need to specify shell=True:
rc = call("./sleep.sh", shell=True)
If the script has no executable permissions and you can’t change it e.g., by running os.chmod('sleep.sh', 0o755) then you could read the script as a text file and pass the string to subprocess module instead:
with open('sleep.sh', 'rb') as file:
script = file.read()
rc = call(script, shell=True)
Adding an answer because I was directed here after asking how to run a bash script from python. You receive an error OSError: [Errno 2] file not found if your script takes in parameters. Lets say for instance your script took in a sleep time parameter: subprocess.call("sleep.sh 10") will not work, you must pass it as an array: subprocess.call(["sleep.sh", 10])
If someone looking for calling a script with arguments
import subprocess
val = subprocess.check_call("./script.sh '%s'" % arg, shell=True)
Remember to convert the args to string before passing, using str(arg).
This can be used to pass as many arguments as desired:
subprocess.check_call("./script.ksh %s %s %s" % (arg1, str(arg2), arg3), shell=True)
If chmod is not working then you can also try:
import os
os.system('sh script.sh')
# you can also use bash instead of sh