[awk] Printing with sed or awk a line following a matching pattern

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.

If pattern match, copy next line into the pattern buffer, delete a return, then quit -- side effect is to print.

sed '/pattern/ { N; s/.*\n//; q }; d'

awk Version:

awk '/regexp/ { getline; print $0; }' filetosearch

Piping some greps can do it (it runs in POSIX shell and under BusyBox):

cat my-file | grep -A1 my-regexp | grep -v -- '--' | grep -v my-regexp

1. -v will show non-matching lines
2. -- is printed by grep to separate each match, so we skip that too

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.

It's the line after that match that you're interesting in, right? In sed, that could be accomplished like so:

sed -n '/ABC/{n;p}' infile

Alternatively, grep's A option might be what you're looking for.

-A NUM, Print NUM lines of trailing context after matching lines.

For example, given the following input file:

foo
bar
baz
bash
bongo

You could use the following:

$ grep -A 1 "bar" file
bar
baz
$ sed -n '/bar/{n;p}' file
baz

Hope that helps.

This might work for you (GNU sed):

sed -n ':a;/regexp/{n;h;p;x;ba}' file

Use seds grep-like option -n and if the current line contains the required regexp replace the current line with the next, copy that line to the hold space (HS), print the line, swap the pattern space (PS) for the HS and repeat.

I needed to print ALL lines after the pattern ( ok Ed, REGEX ), so I settled on this one:

sed -n '/pattern/,$p' # prints all lines after ( and including ) the pattern

But since I wanted to print all the lines AFTER ( and exclude the pattern )

sed -n '/pattern/,$p' | tail -n+2  # all lines after first occurrence of pattern

I suppose in your case you can add a head -1 at the end

sed -n '/pattern/,$p' | tail -n+2 | head -1 # prints line after pattern

Actually sed -n '/pattern/{n;p}' filename will fail if the pattern match continuous lines:

$ seq 15 |sed -n '/1/{n;p}'
2
11
13
15

The expected answers should be:

2
11
12
13
14
15

My solution is:

$ sed -n -r 'x;/_/{x;p;x};x;/pattern/!s/.*//;/pattern/s/.*/_/;h' filename

For example:

$ seq 15 |sed -n -r 'x;/_/{x;p;x};x;/1/!s/.*//;/1/s/.*/_/;h'
2
11
12
13
14
15

Explains:

1. x;: at the beginning of each line from input, use x command to exchange the contents in pattern space & hold space.
2. /_/{x;p;x};: if pattern space, which is the hold space actually, contains _ (this is just a indicator indicating if last line matched the pattern or not), then use x to exchange the actual content of current line to pattern space, use p to print current line, and x to recover this operation.
3. x: recover the contents in pattern space and hold space.
4. /pattern/!s/.*//: if current line does NOT match pattern, which means we should NOT print the NEXT following line, then use s/.*// command to delete all contents in pattern space.
5. /pattern/s/.*/_/: if current line matches pattern, which means we should print the NEXT following line, then we need to set a indicator to tell sed to print NEXT line, so use s/.*/_/ to substitute all contents in pattern space to a _(the second command will use it to judge if last line matched the pattern or not).
6. h: overwrite the hold space with the contents in pattern space; then, the content in hold space is ^_$ which means current line matches the pattern, or ^$, which means current line does NOT match the pattern.
7. the fifth step and sixth step can NOT exchange, because after s/.*/_/, the pattern space can NOT match /pattern/, so the s/.*// MUST be executed!