shell scripting checking python version

Question

I have the following shell script to query the python version. It gives me an error stating "Integer expression expected" on the if-statement.

#!/bin/bash


PYTHON_VERSION=`python -c 'import sys; print("%i" % (sys.hexversion<0x03000000))'`

echo $PYTHON_VERSION

if [ $PYTHON_VERSION -eq 0 ]
then
    echo "fine!"
fi

'echo $PYTHON_VERSION' prints out '0', so why dosn't the if-statement work?

EDIT: I am using Windows and Cygwin

Answer 1

Good question. For me it's working fine. You always should quote evaluated variables ("$X" instead of $X); maybe that fixes your error.

But I propose to use the result of the python script instead of its output:

#!/bin/bash
if python -c 'import sys; sys.exit(1 if sys.hexversion<0x03000000 else 0)'
then
    echo "Fine!"
fi

If you like to stay in the shell completely, you also can use the --version option:

case "$(python --version 2>&1)" in
    *" 3."*)
        echo "Fine!"
        ;;
    *)
        echo "Wrong Python version!"
        ;;
esac

Maybe that's more readable.

Answer 2

The reason why it doesn't work is because the result stored in $PYTHON_VERSION is not an integer, so your equality test is being done with two different types.

You can change the if to:

if [ $PYTHON_VERSION -eq "0" ]; then
     echo "fine!"
fi

or you can just do:

if [ $PYTHON_VERSION = 0 ]; then

Answer 3

here is a possible solution

if sys.version_info < (3, 0):
    reload(sys)
    sys.setdefaultencoding('utf8')
else:
    raw_input = input

this is from an example I had from sleekxmpp, but it does what you need I believe.