printing lines before and after a record

Shell Programming and Scripting
1

Hello Everyone,

I want to print out the records after and before a certain record. I am able to figure out how to print that particular record but not the ones before and after. Looking for some advice

Thank you

2

What OS are you on? Some versions of grep support:

grep -A 1 -B 1 [pattern]  filename

give you one line before and one line after. Is this what you want?

3

thanks for the reply jim

im using hp ux 11.3 and the sad part is, i think, grep doesnt support the linux options :frowning:

4 
 
sed -n -e '/pattern/{x;1!p;g;$!N;p;D;}' -e h infile
5

I don't have a pattern..just a condition.

for eg;

when so and so is true then print the lines above and below it..

6 
 
awk -f d.awk infile

where d.awk:

 
#/bin/ksh
{ if ($2 == 105) {
    print lr;
    print $0;
    lr=$0;
    getline;
    print $0;
  } else {
    lr=$0;
  }
}

The example condition in the script is $2==105.

7

If rdrtx1's suggestion isn't sufficient, stop being vague and specify your condition.

Regards,
Alister