Bash: Extract variable value from string

For a perl-compatible regular expression, you're looking for a "lookbehind" assertion.

To find digits that are preceded by the string "error=", you want:

echo "$line" | grep -o -P '(?<=error=)\d+'    # => 15

See the pcresyntax(3) man page

I would use GNU AWK following way, let file.txt content be

WARN (Periodic Recovery) IJ000906: error=15 check server.log

then

awk 'BEGIN{FPAT="error=[0-9]+"}{print substr($1,7)}' file.txt

output

15

Explanation: I inform GNU AWK that column is error= followed by 1 or more digits using field pattern (FPAT), for every line print first field starting from 7th charater, using substr string function. 7 as error= has 6 characters. Note: this solution will print first occurence of error=value for each line.

(tested in gawk 4.2.1)

You may use this grep:

s='WARN (Periodic Recovery) IJ000906: error=15 check server.log'
grep -oP '\berror=\K\d+' <<< "$s"

15

RegEx Details:

• \b: Match word boundary
• error=: Match error= text
• \K: Reset matched info
• \d+: Match 1+ digits and print it

With bash >= 3.0.

v='WARN (Periodic Recovery) IJ000906: error=15 check server.log'

[[ $v =~ error=([0-9]+) ]] && echo "${BASH_REMATCH[1]}"

Output:

15

1st solution: With your shown samples, please try following awk code.

awk -F'error=| check' '{print $2}' Input_file

Explanation: Simple explanation would be, setting field separators as error= OR check for all the lines. Then printing 2nd field of line, which will print value after error= and before check as per shown samples.

2nd solution: Using match function of awk here.

awk 'match($0,/error=[^[:space:]]+/){print substr($0,RSTART+6,RLENGTH-6)}' Input_file