shell script to find the nth occurrence of a string and print the line number

Question

I have a file named hello with the following data

onefish
onechicken
twofish
twochicken
threechicken
twocows

I want to get the line number at which chicken is occurring for the second time.

Output should be "4" as chicken occurs for the second time in 4th line.

Answer 1

You can use awk for this:

awk '/chicken/{++n; if (n==2) { print NR; exit}}' file

4

Answer 2

grep -n chicken hello| sed -n '2 s/:.*//p'

Answer 3

for fun, using pattern as separtor and counting on field

awk -F 'chicken' 'NF>1{if (2==++o)print NR}' YourFile

using sed (not appropriate compare to awk) [ ^J is CTRL+V+J, real new line]

sed -e '/chicken/H;g;s/\(\^J.*\)\{2\}//;t h' -e 'd' -e ':h' -e ' =;q' YourFile

Answer 4

shorter

$ awk '/chicken/ && ++n==2 {print NR}' file

if the input file is long, adding exit after printing will stop processing.