Count overlapping occurences of a repeated string using grep/linux/bash -

i'm trying count occurences of repeated string. eg.

echo 'joebobtomtomtomjoebobmike' | grep -o 'tomtom' | wc -l 

this outputs 1, string 'tomtom' fits twice here. how can make counts both occurences?

thanks!

you can use awk script

{     count = 0     $0 = tolower($0)     while (length() > 0) {         m = match($0, pattern)         if (m == 0)              break         count++         $0 = substr($0, m + 1)     }     print count } 

explanation

we first convert line lower case ignore case. script works shortening string after matching pattern. uses function match() find position pattern matched. if m == 0, means no matches found, can break loop. increment count each iteration of loop, reset $0 string substring starting @ index m + 1.

if save a.awk, can do

echo "joebobtomtomtomjoebobmike" | awk -v "pattern=tomtom" -f a.awk  

and output 2.