shebang env preferred python version
Question:
I have some python-2.x scripts which I copy between different systems, Debian and Arch linux.
Debian install python as ‘/usr/bin/python’ while Arch installs it as ‘/usr/bin/python2’.
A problem is that on Arch linux ‘/usr/bin/python’ also exists which refers to python-3.x.
So every time I copy a file I have to correct the shebang line, which is a bit annoying.
On Arch I use
#!/usr/bin/env python2
While on debian I have
#!/usr/bin/env python
Since ‘python2’ does not exist on Debian, is there a way to pass a preferred application? Maybe with some shell expansion? I don’t mind if it depends on ‘/bin/sh’ existing for example.
The following would be nice but don’t work.
#!/usr/bin/env python2 python
#!/usr/bin/env python{2,}
#!/bin/sh python{2,}
#!/bin/sh -c python{2,}
The frustrating thing is that ‘sh -c python{2,}’ works on the command line: i.e. it calls python2 where available and otherwise python.
I would prefer not to make a make a link ‘python2->python’ on Debian because then if I give the script to someone else it will not run. Neither would I like to make ‘python’ point to python2 on Arch, since it breaks with updates.
Is there a clean way to do this without writing a wrapper?
I realize similar question have been asked before, but I didn’t see any answers meeting my boundary conditions 🙂
Conditional shebang line for different versions of Python
— UPDATE
I hacked together an ugly shell solution, which does the job for now.
#!/bin/bash
pfound=false; v0=2; v1=6
for p in /{usr/,}bin/python*; do
v=($(python -V 2>&1 | cut -c 7- | sed 's/./ /g'))
if [[ ${v[0]} -eq $v0 && ${v[1]} -eq $v1 ]]; then pfound=true; break; fi
done
if ! $pfound; then echo "no suitable python version (2.6.x) found."; exit 1; fi
$p - $* <<EOF
PYTHON SCRIPT GOES HERE
EOF
explanation:
get version number (v is a bash array) and check
v=($(python -V 2>&1 | cut -c 7- | sed 's/./ /g'))
if [[ ${v[0]} -eq $v0 && ${v[1]} -eq $v1 ]]; then pfound=true; break; fi
launch found program $p with input from stdin (-) and pass arguments ($*)
$p - $* <<EOF
...
EOF
Answers:
Something like this:
#!/usr/bin/env python
import sys
import os
if sys.version_info >= (3, 0):
os.execvp("python2.7", ["python2.7", __file__])
os.execvp("python2.6", ["python2.6", __file__])
os.execvp("python2", ["python2", __file__])
print ("No sutable version of Python found")
exit(2)
Update Below is a more robust version of the same.
#!/bin/bash
ok=bad
for pyth in python python2.7 python2.6 python2; do
pypath=$(type -P $pyth)
if [[ -x $pypath ]] ; then
ok=$(
$pyth <<@@
import sys
if sys.version_info < (3, 0):
print ("ok")
else:
print("bad")
@@
)
if [[ $ok == ok ]] ; then
break
fi
fi
done
if [[ $ok != ok ]]; then
echo "Could not find suitable python version"
exit 2
fi
$pyth <<@@
<<< your python script goes here >>>
@@
#!/bin/sh
''''which python2 >/dev/null 2>&1 && exec python2 "$0" "$@" # '''
''''which python >/dev/null 2>&1 && exec python "$0" "$@" # '''
''''exec echo "Error: I can't find python anywhere" # '''
import sys
print sys.argv
This is first run as a shell script. You can put almost any shell code in between '''' and # '''. Such code will be executed by the shell. Then, when python runs on the file, python will ignore the lines as they look like triple-quoted strings to python.
The shell script tests if the binary exists in the path with which python2 >/dev/null and then executes it if so (with all arguments in the right place). For more on this, see Why does this snippet with a shebang #!/bin/sh and exec python inside 4 single quotes work?
Note: The line starts with four ' and their must be no space between the fourth ' and the start of the shell command (which…)
I’ll leave this here for future reference.
All of my own scripts are usually written for Python 3, so I’m using a modified version of Aaron McDaid’s answer to check for Python 3 instead of 2:
#!/usr/bin/env sh
''''which python3 >/dev/null 2>&1 && exec python3 "$0" "$@" # '''
''''test $(python --version 2>&1 | cut -c 8) -eq 3 && exec python "$0" "$@" # '''
''''exec echo "Python 3 not found." # '''
import sys
print sys.argv
Here is a more concise way for the highest voted answer:
#!/bin/sh
''''exec $(which python3 || which python2 || echo python) $0 $@ #'''
import sys
print(sys.argv)
print(sys.version_info)
You’ll get this if none of them found:
'./test.py: 2: exec: python: not found'
Also, get rid of warnings from linter:
'module level import not at top of file - flake8(E402)'
In my case I don’t need the python path or other information (like version).
I have an alias setup to point "python" to "python3". But, my scripts are not aware of the environment alias. Therefore I needed a solution for the scripts to determine the program name, programmatically.
If you have more than one version installed, the sort and tail are going to give you the latest version:
#!/bin/sh
command=$((which python3 || which python2 || which python) | sort | tail -n1 | awk -F "/" '{ print $NF }')
echo $command
Would give a result like: python3
This solution is original but similar to @Pamela
I have some python-2.x scripts which I copy between different systems, Debian and Arch linux.
Debian install python as ‘/usr/bin/python’ while Arch installs it as ‘/usr/bin/python2’.
A problem is that on Arch linux ‘/usr/bin/python’ also exists which refers to python-3.x.
So every time I copy a file I have to correct the shebang line, which is a bit annoying.
On Arch I use
#!/usr/bin/env python2
While on debian I have
#!/usr/bin/env python
Since ‘python2’ does not exist on Debian, is there a way to pass a preferred application? Maybe with some shell expansion? I don’t mind if it depends on ‘/bin/sh’ existing for example.
The following would be nice but don’t work.
#!/usr/bin/env python2 python
#!/usr/bin/env python{2,}
#!/bin/sh python{2,}
#!/bin/sh -c python{2,}
The frustrating thing is that ‘sh -c python{2,}’ works on the command line: i.e. it calls python2 where available and otherwise python.
I would prefer not to make a make a link ‘python2->python’ on Debian because then if I give the script to someone else it will not run. Neither would I like to make ‘python’ point to python2 on Arch, since it breaks with updates.
Is there a clean way to do this without writing a wrapper?
I realize similar question have been asked before, but I didn’t see any answers meeting my boundary conditions 🙂
Conditional shebang line for different versions of Python
— UPDATE
I hacked together an ugly shell solution, which does the job for now.
#!/bin/bash
pfound=false; v0=2; v1=6
for p in /{usr/,}bin/python*; do
v=($(python -V 2>&1 | cut -c 7- | sed 's/./ /g'))
if [[ ${v[0]} -eq $v0 && ${v[1]} -eq $v1 ]]; then pfound=true; break; fi
done
if ! $pfound; then echo "no suitable python version (2.6.x) found."; exit 1; fi
$p - $* <<EOF
PYTHON SCRIPT GOES HERE
EOF
explanation:
get version number (v is a bash array) and check
v=($(python -V 2>&1 | cut -c 7- | sed 's/./ /g'))
if [[ ${v[0]} -eq $v0 && ${v[1]} -eq $v1 ]]; then pfound=true; break; fi
launch found program $p with input from stdin (-) and pass arguments ($*)
$p - $* <<EOF
...
EOF
Something like this:
#!/usr/bin/env python
import sys
import os
if sys.version_info >= (3, 0):
os.execvp("python2.7", ["python2.7", __file__])
os.execvp("python2.6", ["python2.6", __file__])
os.execvp("python2", ["python2", __file__])
print ("No sutable version of Python found")
exit(2)
Update Below is a more robust version of the same.
#!/bin/bash
ok=bad
for pyth in python python2.7 python2.6 python2; do
pypath=$(type -P $pyth)
if [[ -x $pypath ]] ; then
ok=$(
$pyth <<@@
import sys
if sys.version_info < (3, 0):
print ("ok")
else:
print("bad")
@@
)
if [[ $ok == ok ]] ; then
break
fi
fi
done
if [[ $ok != ok ]]; then
echo "Could not find suitable python version"
exit 2
fi
$pyth <<@@
<<< your python script goes here >>>
@@
#!/bin/sh
''''which python2 >/dev/null 2>&1 && exec python2 "$0" "$@" # '''
''''which python >/dev/null 2>&1 && exec python "$0" "$@" # '''
''''exec echo "Error: I can't find python anywhere" # '''
import sys
print sys.argv
This is first run as a shell script. You can put almost any shell code in between '''' and # '''. Such code will be executed by the shell. Then, when python runs on the file, python will ignore the lines as they look like triple-quoted strings to python.
The shell script tests if the binary exists in the path with which python2 >/dev/null and then executes it if so (with all arguments in the right place). For more on this, see Why does this snippet with a shebang #!/bin/sh and exec python inside 4 single quotes work?
Note: The line starts with four ' and their must be no space between the fourth ' and the start of the shell command (which…)
I’ll leave this here for future reference.
All of my own scripts are usually written for Python 3, so I’m using a modified version of Aaron McDaid’s answer to check for Python 3 instead of 2:
#!/usr/bin/env sh
''''which python3 >/dev/null 2>&1 && exec python3 "$0" "$@" # '''
''''test $(python --version 2>&1 | cut -c 8) -eq 3 && exec python "$0" "$@" # '''
''''exec echo "Python 3 not found." # '''
import sys
print sys.argv
Here is a more concise way for the highest voted answer:
#!/bin/sh
''''exec $(which python3 || which python2 || echo python) $0 $@ #'''
import sys
print(sys.argv)
print(sys.version_info)
You’ll get this if none of them found:
'./test.py: 2: exec: python: not found'
Also, get rid of warnings from linter:
'module level import not at top of file - flake8(E402)'
In my case I don’t need the python path or other information (like version).
I have an alias setup to point "python" to "python3". But, my scripts are not aware of the environment alias. Therefore I needed a solution for the scripts to determine the program name, programmatically.
If you have more than one version installed, the sort and tail are going to give you the latest version:
#!/bin/sh
command=$((which python3 || which python2 || which python) | sort | tail -n1 | awk -F "/" '{ print $NF }')
echo $command
Would give a result like: python3
This solution is original but similar to @Pamela