Sorting not working correctly in bash

Question

I have a variable VarExp with following 2 values

1.5.2
1.5.3

I have another variable VarCurr with following 1 value

1.8.1

I want to compare VarCurr with VarExp and want to echo SUCCESS only when

VarCurr >= VarExp

I have written the following code but it it always returning FAILURE

VarExp='1.5.2 1.5.3'
VarCurr='1.8.1'

printf -v versions '%s\n%s' "$VarExp" "$VarCurr"
if [[ $versions = "$(sort -V <<< "$versions")" ]]; then
    echo 'FAILURE'
else
    echo 'SUCCESS'
fi

VarCurr need to be >= the lowest value contained in VarExp

Answer 1

I recommend using a language that can properly objectify version objects and can understand major.minor.build.revision. Here's an example bash script which borrows from Perl for version parsing:

#!/bin/bash

VarExp='1.5.2 1.5.3'
VarCurr='1.8.1'

for i in $VarExp; do {
    perl -e 'use version;exit !(version->parse('$VarCurr') >= version->parse('$i'));' && {
        echo 'SUCCESS'
        exit
    }
}; done

echo 'FAILURE'
exit

Of course it might be more graceful just to write the whole thing in Perl.

Edit: Here's another example using Python:

#!/bin/bash

VarExp='1.5.3 1.5.6'
VarCurr='1.5.3'

for i in $VarExp; do {
    python -c 'from distutils.version import LooseVersion;\
    exit(LooseVersion("'$VarCurr'") >= LooseVersion("'$i'"))' || {
        echo 'SUCCESS'
        exit
    }
}; done

echo 'FAILURE'
exit

Answer 2

With GNU sort for -V:

$ cat tst.sh
#!/bin/bash
varExp='1.5.2 1.5.3'
varCurr=$1

minVarExp=$(printf '%s\n' $varExp | sort -V | head -1)
maxOfVers=$(printf '%s\n' "$minVarExp" "$varCurr" | sort -V | tail -1)

if [[ $maxOfVers = $varCurr ]]; then
    echo 'SUCCESS'
else
    echo 'FAILURE'
fi

$ ./tst.sh 1.8.1
SUCCESS

$ ./tst.sh 1.5.1
FAILURE

$ ./tst.sh 1.5.2
SUCCESS