I tried the following solution.
if !(($str =~ ^.*[a-zA-Z0-9].*$)); then
continue;
fi
It seems there's something wrong with the syntax.
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2 Answers
Both your regex and syntax is incorrect. You need not use regex at all, just need glob for this:
checkStr() {
[[ $1 != *[a-zA-Z0-9]* ]]
}
This will return false if even one alphanumeric is found in the input. Otherwise true is returned.
Test it:
$> if checkStr 'abc'; then
echo "true case"
else
echo "false case"
fi
false case
$> if checkStr '=:() '; then
echo "true case"
else
echo "false case"
fi
true case
$> if checkStr '=:(x) '; then
echo "true case"
else
echo "false case"
fi
false case
3 Comments
xralf
I need the expression that returns True if a string matches a mix of zero or more whitespace and zero or more non-alphanumeric characters. If there is at least one alpha-numeric character it should return False. That's because I find the opposite and then negate the result with !
chepner
@xralf The
!= operator here does that, unless you have a specific set of nonalphanumeric characters in mind.xralf
@anubhava The examples you provided are exactly what I thought.
