Print Line above and line below

Shell Programming and Scripting
1

Hello evryone,

I am looking for a way to print x line above and x line below ...

The idea is : I've got a little script that I am trying to make it universal to read debug log file.

I search for specific error value ( ERROR, Failed, NOK) and so on..., but most of the time more info can be useful above and below the line printed.

I am lookink for a way, in interative script to select value above and below the line printed but I didn't find any post to help me on this.

Like print 2 lines above and 3 lines below the line printed with the word ERROR for exemple ....

I hope, someone will be able to help me. Please I can only user ksh. I can't use Perl.

Thanks in advanced for your help.

2 
grep -A3 -B2 ERROR file.log
1 Like
3

If your grep version doesn't support the -A or -B option you could try something like:

awk '/ERROR/{print s RS $0; getline; print; next}{s=$0}' file
1 Like
4

Thanks Franklin52 for your help.

I think I will go for your code, but I didn't see where I could configure how many lines above and below ?

But I also have trouble as where I work the log file are huge, it's unbelievable ... I need to be able to filter by date, but I can't see where I could add this varibale on your line code ?

Sorry to not be good enough on this ...

Thanks for your help

5

Maybe some tool designed to log analysis could help your case? Check Download Splunk for free on your operating system

6

Thanks Bartus11 for your help, as I can't install anything on the server, I can't try your tool.

Anyway, I will try to work on the codes given, and see ....

Thanks.

7

Here is an example how to print 3 lines above and 4 lines after the pattern:

awk '$0 ~ s {
  for(i=0;i<b;i++){print arr[(NR+i)%b]}
  for(i=0;i<a;i++){print;getline}
  print;exit
}
{arr[NR%b]=$0}
' b=3 a=4 s="ERROR" file
8

How about plain old ex...this will print 4 lines above/below "ERROR"...

ex -s +'/ERROR/-4,/ERROR/+4 p | q!' file
1 Like