Printing with sed or awk a line following a matching pattern

Question

Question: I'd like to print a single line directly following a line that contains a matching pattern.

My version of sed will not take the following syntax (it bombs out on +1p) which would seem like a simple solution:

sed -n '/ABC/,+1p' infile

I assume awk would be better to do multiline processing, but I am not sure how to do it.

Answer 1

Never use the word "pattern" as is it highly ambiguous. Always use "string" or "regexp" (or in shell "globbing pattern"), whichever it is you really mean.

The specific answer you want is:

awk 'f{print;f=0} /regexp/{f=1}' file

or specializing the more general solution of the Nth record after a regexp (idiom "c" below):

awk 'c&&!--c; /regexp/{c=1}' file

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

a) Print all records from some regexp:

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

b) Print all records after some regexp:

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

c) Print the Nth record after some regexp:

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

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

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

e) Print the N records after some regexp:

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

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

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

g) Print the N records from some regexp:

awk '/regexp/{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.