extract a string after a pattern

This may work for you:

awk -F "[):,]" '{ for (i=2; i<=NF; i++) if ($i ~ /id/) print $2, $(i+1) }' file

Results:

03 20403 49103 2904 20904 30105 20

Here's a awk script that works (I put it on multiple lines and made it a bit more verbose so you can see what's going on):

#!/bin/bashawk 'BEGIN{FS="[\(\):,]"}/client_id/ {cid="no_client_id"for (i=1; i<NF; i++) {    if ($i == "client_id") {        cid = $(i+1)    } else if ($i == "id") {        id = $(i+1);        print cid OFS id;    } }}' input_file_name

Output:

03 20403 49103 2904 20904 30105 20

Explanation:

• awk 'BEGIN{FS="[\(\):,]"}: invoke awk, use ( ) : and , as delimiters to separate your fields
• /client_id/ {: Only do the following for the lines that contain client_id:
• for (i=1; i<NF; i++) {: iterate through the fields on each line one field at a time
• if ($i == "client_id") { cid = $(i+1) }: if the field we are currently on is client_id, then its value is the next field in order.
• else if ($i == "id") { id = $(i+1); print cid OFS id;}: otherwise if the field we are currently on is id, then print the client_id : id pair onto stdout
• input_file_name: supply the name of your input file as first argument to the awk script.

This might work for you (GNU sed):

sed -r '/.*(\(client_id:([0-9]+))[^(]*\(id:([0-9]+)/!d;s//\2 \3\n\1/;P;D' file

• /.*(\(client_id:([0-9]+))[^(]*\(id:([0-9]+)/!d if the line doesn't have the intended strings delete it.
• s//\2 \3\n\1/ re-arrange the line by copying the client_id and moving the first id ahead thus reducing the line for successive iterations.
• P print upto the introduced newline.
• D delete upto the introduced newline.