How to iteratively find number of lines between two patterns?

Question

This question is similar, but distinct from 1, 2, and 3.

Instead of pasting the lines in between the two patterns, I want to iteratively count the lines.

For example, given file.txt with these strings

abc
123
daafsd
asdfas
asdcasdfa
123
sdfasdc
asdfasdcasd
asdfasdfasdf
asdfasdfasdf
ascasdcasdcasd
123
asdcasdfacasdcas
123
asdfasdcasdcasc
asadfasdfas
123

I would want to count the lines between the pattern of 123. So, the expected output would be:

3
5
1
2

Any suggestions?

Answer 1

awk '$0=="123" {if (n) print NR-1-n; n=NR}' file

This uses the line number of matched lines to print the number of lines between them.

Answer 2

With your shown samples/attempts, please try following awk code.

awk -v RS='(^|\n)123(\n|$)' 'FNR>1 && NF{print NF}' Input_file

For shown samples, output will be as follows:

3
5
1
2

Explanation:

• In awk program setting RS(record separator) as (^|\n)123(\n|$).
• Explanation of regex (^|\n)123(\n|$): Setting record separator as 123(which can come at starting of line OR could be followed or start with a new line OR end with 123), as per shown samples.
• Then in main program checking condition if line number is greater than 1 AND NF(number of fields) is NOT zero then print the NF(total number of fields) for that line, which will be number of lines came between patterns.

Answer 3

This might work for you (GNU sed and wc):

sed -En ':a;/123/{:b;n;//!{H;bb};x;s/.(.*)/echo "\1"|wc -l/ep;z;x;ba}' file

Turn on extended regexp -E and off implicit printing -n.

Match on 123, fetch the next line, if it does not match, store the result in the hold space and repeat.

Within a substitution command, evaluate the result of the tail of the lines and count them using wc -l.

Clear the hold space, swap back to the pattern space and repeat.