Ygtjki

I need to solve this in a shell script. I am counting number of occurrence of the abc string below and I want to get the answer as 3.

echo abcsdabcsdabc | grep -o abc

abc

abc

abc

Assuming we do not have the -o option in grep, how do we approach this then?

With awk:

awk 'BEGINprint gsub(ARGV[2], "&", ARGV[1])' abcsdabcsdabc abc

Note that the pattern (here abc) is taken by awk as an extended regular expression (as if using grep -E/egrep).

That syntax allows both the subject and regexp to contain multiple lines. We also avoid the usual problems associated with echo which can't output arbitrary data.

To use perl regular expressions (similar to GNU grep -P's):

perl -le 'print scalar (() = $ARGV[0] =~ m$ARGV[1]g)' -- abcsdabcsdabc abc

(note however that the arguments are not interpreted as text as per the locale's encoding. For instance in a UTF-8 locale, with é and . as arguments, it would report 2 (bytes) instead of 1 (character)).

Treating the string as consisting of fields that are delimited by abc:

$ echo abcsdabcsdabc | awk -F 'abc' ' print NF - 1 '
3

The number of occurrences of the delimiter abc is 1 minus the number of fields that it delimits.

$ echo abcsdabcsdabc | awk ' n=0; while (sub("abc", "")) n++; print n '
3

This deletes the substring abc from the line and counts the number of times this is done, then outputs that number. The n=0 is not needed if there is only one line of input.

The gsub() function in awk returns the number of substitutions made, so the above could be simplified into

$ echo abcsdabcsdabc | awk ' print gsub("abc", "") '
3

In bash, you can do the same thing as in that awk program that uses sub():

string=abcsdabcsdabc

n=0
while [[ $string == *abc* ]]; do
 n=$(( n+1 ))
 string=$string/abc/ # remove first occurrence of "abc"
done
printf '%dn' "$n"

This uses a while loop to remove the substring abc from the value in $string until no further occurrences of abc is found in $string, just as the second awk program above does.