Sed replace at second occurrence

Question

I want to remove a pattern with sed, only at second occurence. Here is what I want, remove a pattern but on second occurrence.

What's in the file.csv:

a,Name(null)abc.csv,c,d,Name(null)abc.csv,f
a,Name(null)acb.csv,c,d,Name(null)acb.csv,f
a,Name(null)cba.csv,c,d,Name(null)cba.csv,f

Output wanted:

a,Name(null)abc.csv,c,d,Name,f
a,Name(null)acb.csv,c,d,Name,f
a,Name(null)cba.csv,c,d,Name,f

This is what i tried:

sed -r 's/(\(null)\).*csv//' file.csv

The problem here is that the regex is too greedy, but i cannot make is stop. I also tried this, to skip the first occurrence of "null":

sed -r '0,/null/! s/(\(null)\).*csv//' file.csv

Also tried but the greedy regex is still the problem.

sed -r 's/(\(null)\).*csv//2' file.csv

I've read that ? can make the regex "lazy", but I cannot make it workout.

sed -r 's/(\(null)\).*?csv//' file.csv

Answer 1

sed does provide an easy way to specify which match to be replaced. Just add the number after delimiters

$ sed 's/(null)[^.]*\.csv//2' ip.csv
a,Name(null)abc.csv,c,d,Name,f
a,Name(null)acb.csv,c,d,Name,f
a,Name(null)cba.csv,c,d,Name,f

$ # or [^,] if there are no , within fields
$ sed 's/(null)[^,]*//2' ip.csv
a,Name(null)abc.csv,c,d,Name,f
a,Name(null)acb.csv,c,d,Name,f
a,Name(null)cba.csv,c,d,Name,f

Also, no need to escape () when not using extended regular expressions