Shell Programming and Scripting

Greetings,

I am writing a script which requires as a part, selecting all the lines between the first and last occurrence of a pattern.

I have an nawk alternative that is working. But thre should be a generic script that should run on all os viz, linux, sun , aix.

The awk script that i am using is given below.

Awk script

Code:

 
nawk 'NR==FNR{if(/xyz/){last=NR;if(!first)first=NR};next} FNR>=first && FNR<=last' test.txt test.txt

Can anyone help me with the sed substitute for selecting everything between the first and last occurrence of a pattern in just one iteration.

Thanks in advance.
Usha

Code:

 
Is this what you are looking for 
Input:
this is test
one
two
three
four
two
one
end of file

"one" is the search string here.

Code:

Code:

perl -0ne 'print $1 if /one(.*)one/sg' test

Output:

Code:

 
two
three
four
two

Thanks for the reply getmmg.

But I need the lines containg the pattern also to be printed.

sample output from your example:

Code:

 
one
two
three
four
two
one

Also, i do not have perl in my system, Can you suggest some sed code?

Thanks,
Usha Rao

Hello Usha,

How about this?

Code:

awk -v c=0 '/xyz/&&c++<1 {a=$0;next} /xyz/&&c>0 { a=a"\n"$0;f=a;next} c>0 {a=a"\n"$0;next} END { print f}' input_file

Code:

sed -n '/one/,/one/p'

Hi Panyam,

Thanks for the reply.
The given awk solution perfectly works when i am using nawk in my solaris system.

But i am writing a generic script that should be common for all solaris , linux and aix system.

Is there any SED substitute.


Thanks,
Usha Rao

I believe "awk" is every where!!!! In AIX, LINUX and UNIX too!!!..