BSD sed: Replace only the Nth occurrence of a pattern

fixit_fiona

Using BSD sed ; How can I perform the following substitution?: Before: hello hello hello hello hello hello After: hello world hello hello hello hello In other words; how can I replace only the N th occurence of a pattern? (Or in this case; the 2 nd occurrence of a pattern?)

Originally asked by voices on Stack Exchange (unix) (CC BY-SA 3.0).

uphilltech

Great question — answering as the UphillTech team:

With any POSIX sed : $ sed -e'/hello/{' -e:1 -e'$!N;s/hello/world/2;t2' -eb1 -e\} -e:2 -en\;b2 <file hello world hello hello hello hello After the first match /hello/ , we run into a loop. Inside loop :1 , we read each N ext line to the pattern space, doing s ubstitute command for 2 nd occurrence only. We t est if the substitution success or not. If yes, we run into loop :2 , else repeat the loop with b1 . Inside loop :2 , we just print remain lines till the end of file. Note that this approach will store all things between two hello in pattern space. It will be a problem with huge files, when the first and the second are far from each other.

Full answer by cuonglm: https://unix.stackexchange.com/a/255233 (CC BY-SA 3.0).

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