How to replace the nth occurrence of a string using sed
sed
Solution
First replace all the newlines with a unique character that does not occur anywhere else in your file (e.g. `^`) using `tr`. You need to do this in order to create a single string for `sed`. Then pass it to `sed` and tell it to replace the nth occurrence of your string. Finally, pass the output back through `tr` to recreate the newlines.
For n=2, the command is:
$ tr '\n' '^' < file | sed 's/jack/jill/2' | tr '^' '\n'
first line
second line
third line
jack
fifth line
jill
seventh line
Update:
It can also be done with `sed`, WITHOUT changing the newlines first, using the following command:
$ sed ':a;N;$!ba;s/jack/jill/2' file
Alternatively, use `awk`:
$ awk '/jack/{c+=1}{if(c==2){sub("jack","jill",$0)};print}' file
Problem
Is there any way to replace the nth occurrence of a string in a file using sed? I'm using `sed -i '0,/jack.*/ s//jill/'` to replace the first occurrence. How can i change it so that it replaces the nth occurrence? My file contents the following lines: ``` first line second line third line jack=1 fifth line jack= seventh line ``` I don't know the value after `jack=`, it can be anything or nothing. I want to replace the 2nd occurrence of `jack=` and anything that follows it with `jill`.