linux - replace every nth occurrence of a pattern using awk -

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this question has answer here:

i'm trying replace every nth occurrence of string in text file.

background: have huge bibtex file (called in.bib) containing hundreds of entries beginning "@". every entry has different amount of lines. want write string (e.g. "#") right before every (let's say) 6th occurrence of "@" so, in second step, can use csplit split huge file @ "#" files containing 5 entries each.

the problem find , replace every fifth "@".

since need repeatedly, suggested answer in printing sed or awk line following matching pattern won't job. again, not looking 1 matching place many of it.

what have far:

awk '/^@/ && v++%5 {sub(/^@/, "\n#\n@")} {print > "out.bib"}' in.bib

replaces 2nd until 5th occurance (and no more). (btw, found , adopted solution here: "sed replace every nth occurrence". initially, meant replace every second occurence--which does.)

and, second:

awk -v p="@" -v n="5" '$0~p{i++}i==n{sub(/^@/, "\n#\n@")}{print > "out.bib"}' in.bib

replaces 5th occurance , nothing else. (adopted solution here: "display n'th match of grep"

what need (and not able write) imho loop. loop job? like:

for (i = 1; <= 200; * 5)    <find "@"> , <replace "\n#\n@"> print

the material have looks this:

@article{karamanic_jedno_2007,     title = {jedno kosova, dva srbije},     journal = {ulaznica: journal culture, art , social issues},     author = {karamanic, slobodan},     year = {2007} }  @inproceedings{blome_eigene_2008,     title = {das eigene, das andere und ihre vermischung. zur rolle von sexualität und reproduktion im rassendiskurs des 19. jahrhunderts},     comment = {rest of lines snippet off here usability -- in following entries. original entries may have different amount of lines.} }  @book{doring_inter-agency_2008,     title = {inter-agency coordination in united nations peacebuilding} }  @book{reckwitz_subjekt_2008,     address = {bielefeld},     title = {subjekt} }

what want every sixth entry looking this:

# @book{reckwitz_subjekt_2008,     address = {bielefeld},     title = {subjekt} }

thanks help.

your code right, modified it.

to replace every nth occurrence, need modular expression.

so better understanding brackets, need expression ((i % n) == 0) 

awk -v p="@" -v n="5" ' $0~p { i++ } ((i%n)==0) { sub(/^@/, "\n#\n@") }{ print }' in.bib > out.bib

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