Grep a string with number greater than 45

Question

I have multiple files in a directory. I want to extract each line in all the files containing which has integer value greater than 45.

Currently, I am using :

grep "IO resumed after" *

Its displaying me all the files which this string "IO resumed after" I want to put one more parameter that will grep all the lines "IO resumed after [number >45] seconds"

Answer 1

Looks like i need to learn awk until then i've got a bash solution. If seconds without decimal point then this:

while read line; do
    number=${line//*after}
    number=${number//seconds*}
    ((number>45)) && echo $line
done <<< $(grep "IO resumed after" *)

otherwise we have to use bc:

while read line; do
    number=${line//*after}
    number=${number//seconds*}
    case $(bc <<< "$number>45") in 1) echo "$line";; esac
done <<< $(grep "IO resumed after" *)