How to find/replace and increment a matched number with sed/awk?

Question

Straight to the point, I\'m wondering how to use grep/find/sed/awk to match a certain string (that ends with a number) and increment that number by 1. The closest I\'ve come

Answer 1

I think finding file isn't the difficult part for you. I therefore just go to the point, to do the +1 calculation. If you have gnu sed, it could be done in this way:

sed -r 's/(.*)(\?cache_version=)([0-9]+)(.*)/echo "\1\2$((\3+1))\4"/ge' file

let's take an example:

kent$  cat test 
ello
barbaz?cache_version=3fooooo
bye

kent$  sed -r 's/(.*)(\?cache_version=)([0-9]+)(.*)/echo "\1\2$((\3+1))\4"/ge' test     
ello                                                                             
barbaz?cache_version=4fooooo
bye

you could add -i option if you like.

edit

/e allows you to pass matched part to external command, and do substitution with the execution result. Gnu sed only.

see this example: external command/tool echo, bc are used

kent$  echo "result:3*3"|sed -r 's/(result:)(.*)/echo \1$(echo "\2"\|bc)/ge'       

gives output:

result:9

you could use other powerful external command, like cut, sed (again), awk...