Extract a property value from a text file - bash

Question

I have a log file which contains lines like the following one:
Internal (reserved=1728469KB, committed=1728469KB)
I'd need to extract the value contained in "committed", so 1728469
I'm trying to use awk for that
cat file.txt | awk '{print $4}'
However that produces:
committed=1728469KB)
This is still incomplete and would need still some work. Is there a simpler solution to do that instead?
Thanks

Answer 1

Could you please try following, using match function of awk.
awk 'match($0,/committed=[0-9]+/){print substr($0,RSTART+10,RLENGTH-10)}' Input_file
With GNU grep using \K option of it:
grep -oP '.*committed=\K[0-9]*' Input_file
Output will be 1728469 in both above solutions.
1st solution explanation:
awk ' ##Starting awk program from here.
match($0,/committed=[0-9]+/){ ##Using match function to match from committed= till digits in current line.
print substr($0,RSTART+10,RLENGTH-10) ##Printing sub string from RSTART+10 to RLENGTH-10 in current line.
}
' Input_file ##Mentioning Input_file name here.

Answer 2

Sed is better at simple matching tasks:
sed -n 's/.*committed=\([0-9]*\).*/\1/p' input_file

Answer 3

$ awk -F'[=)]' '{print $3}' file
1728469KB

Answer 4

You can try this:
str="Internal (reserved=1728469KB, committed=1728469KB)"
echo $str | awk '{print $3}' | cut -d "=" -f2 | rev | cut -c4- | rev