regex - print specific number of lines after matching pattern

Question

I have to print 81 lines after each occurrence of the expression "AAA" from my input file. How do I go about that?

Answer 1

The following idioms describe how to select a range of records given a specific pattern to match:

a) Print all records from some pattern:

awk '/pattern/{f=1}f' file

b) Print all records after some pattern:

awk 'f;/pattern/{f=1}' file

c) Print the Nth record after some pattern:

awk 'c&&!--c;/pattern/{c=N}' file

d) Print every record except the Nth record after some pattern:

awk 'c&&!--c{next}/pattern/{c=N}1' file

e) Print the N records after some pattern:

awk 'c&&c--;/pattern/{c=N}' file

f) Print every record except the N records after some pattern:

awk 'c&&c--{next}/pattern/{c=N}1' file

g) Print the N records from some pattern:

awk '/pattern/{c=N}c&&c--' file

I changed the variable name from "f" for "found" to "c" for "count" where appropriate as that's more expressive of what the variable actually IS.

So, you'd want "e" above:

awk 'c&&c--;/AAA/{c=81}' file