regex, only the first occurence

Question

I have a pipelined output from free -o -m | awk '{print $4}' I want take step by step two numbers the first and the second line. So i need two regexes. I tried with ^[0-9]{1,3}$ but this match all occurences. i.e. 111 222 333

I want only 111 by first and 222 by second regex.

Answer 1

Your question is a bit backward.

You probably want something like:

free -o -m | awk '
  $1 == "Mem:" {print "do something with " $4}
  $2 == "Swap:" {print "do something with " $4}'

Something like:

eval "$(free -o -m | awk '
  $1 == "Mem:" {print "free_mem=" $4}
  $1 == "Swap:" {print "free_swap=" $4}'
)"
printf "%s\n" "Free mem: $free_mem" "Free swap: $free_swap"

Answer 2

I think I still don't understand completely what you're asking here. But free -o -m | awk '{print $4}' gives you output of the form:

shared
123
456

To check 123 and 456 against a condition on bash you could use a for loop instead of fiddeling with regexes:

for value in `free -o -m | awk '{print $4}'`; do 
    if [[ "$value" != "shared" && CONDITION ]]; then echo "$value matched my condition"; fi; 
done;