awk with $var as search pattern

Question

I am facing a problem where I want awk to search for a pattern that comes from a bash variable... Now it looks like:

for i in ${CLASS_LIST[*]}; do
cat log.txt | awk -F ":" '$1 ~ /$i/ {print substr($0, index($0,$2))}'
done

But awk has no output here. If I do it like:

cat log.txt | awk -F ":" '$1 ~ /LABEL/ {print substr($0, index($0,$2))}

it works well... what is wrong in the first example?

Thanks a lot for your help in advance. regards, Joerg

See: [Difference between single and double quotes in bash](http://stackoverflow.com/q/6697753/3776858) — Cyrus, Oct 27 '20 at 11:49
Take a look at awk's option `-v`. See: [Can we use shell variables in awk?](https://stackoverflow.com/q/15786777/3776858) — Cyrus, Oct 27 '20 at 11:50

score 0 · Accepted Answer · answered Oct 27 '20 at 12:15

Thanks for the tip, Cyrus. I've already tried the "-v" option with the same result. But I found my mistake...

when give a variable as search pattern, don't pack it in to "/"... This is working for me:

for i in ${CLASS_LIST[*]}; do
cat log.txt | awk -v var="$i" -F ":" '$1 ~ var {print substr($0, index($0,$2))}'
done

Regards, Joerg

1 Answers1