[SOLVED] How can I use GREP to decide if every line in a file matches my pattern?

Issue

I am trying to check if every line in a file matches my pattern (4 characters followed by 4 digits). I tried using GREP with -x -P -v -q options so it returns 1 if my file doesn't match the requirements. I expect it to return nothing in case the file is correct, but it returns nothing even if the file has an error. $4 is my input file. My code is:

if [ -f $4 ] && [ `grep -q -P -x -v [[a-z]x{4}/[\dx{4}] $4` ]
then
echo "error"
exit 1
fi

input example: bmkj2132 ahgc3478 (no uppercase)

Solution

Don't check the output of grep, check its exit code.

if [ -f "$4" ] && grep -q ...

Note the double quotes around $4 - otherwise a file name containing whitespace will break the script.

Also, single quote the regex. Square brackets are special in bash and you don't want the regex to suddenly change (expand) when a random filename exists.

Also note that x doesn't mean "times" (under -e, -E, -P neither). The quantifier just follows the quantified with no operator in between:

echo abcd1234 | grep -xP '[a-z]{4}\d{4}'

So, the full condition should be

if [ -f "$4" ] && grep -qvxP '[a-z]{4}\d{4}' "$4" ; then
    echo Error >&2
    exit 1
fi

Are you sure you want to continue if the file doesn't exist? If not, change the condition to

if ! [ -f "$4" ] || grep ...

BTW, you don't really need the PCRE expression here. If you replace \d by [0-9] or [[:digit:]], you can switch to -E (extended regular expression).

Answered By - choroba