CODEWITHSUNDEEP
sed
pfnuesel
pfnuesel

Reputation: 15340

Using regex for address range

I have the following file:

$ cat file
> First line
> Second line
> Third line
> Fourth and last line
> First line
> Second line
> Third line
> Fourth and last line

I want to print the first 3 lines, easy:

$ sed -n '1,3p' file
> First line
> Second line
> Third line

Now I want to print from the occurrence First to the occurrence Third:

$ sed -n '/First/,/Third/p' file
> First line
> Second line
> Third line
> First line
> Second line
> Third line

Ah! Not quite what I wanted, I want only the first occurrence of the matched pattern range. How can I do that when I have regex' as my addresses?

Upvotes: 1

Views: 160

Answers (4)

Thor
Thor

Reputation: 47189

Append the end-pattern as a quit condition:

sed -n '/First/,/Third/p; /Third/q' file

Output:

> First line
> Second line
> Third line

Upvotes: 3

Jotne
Jotne

Reputation: 41460

You can do like this with awk

awk '!f; /Third/ {f=1}' file
> First line
> Second line
> Third line

Or even shorter and better, since it stops processing file after found.

awk '1; /Third/ {exit}' file

Or if its needed to take from first to third

awk '/First/ {f=1} f; /Third/ {exit}' file

Upvotes: 2

sam
sam

Reputation: 2486

I believe awk may help you to do this 

  awk '/First/{found=1} found{print; if(/Third/) exit}' file

Upvotes: 2

sat
sat

Reputation: 14949

You can try this sed:

sed -n '/First/{:loop; $!N; /Third/{p;q}; b loop;}' file

Upvotes: 2

Related Questions