awk command to retrieve record 23 and 89 from UNIX file

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1. The problem statement, all variables and given/known data:

I am looking for awk command to retrieve only the record number 23 and record number 89 from a unix file? Please let me know what is the awk command for this?

2. Relevant commands, code, scripts, algorithms:

Awk command in Unix

3. The attempts at a solution (include all code and scripts):

Looking for only awk command only and not for other commands like sed , head or tail.

4. Complete Name of School (University), City (State), Country, Name of Professor, and Course Number (Link to Course):

I am looking for a job on where it needs Unix skill.So I am exploring the possible opportunities to gain the skill. I completed my University 2 years ago.I am not studying in any school or university now:b:

Note: Without school/professor/course information, you will be banned if you post here! You must complete the entire template (not just parts of it).

So it's actually not home work for a course...

Did you read the man pages for awk?

man awk

The record number (usually the line number) is NR. Also because the main code is a loop, you can simply increment an own counter variable.

Hi

yes I studied and know that how to retrieve only one record with the below command awk 'NR==23' emp.lst but with the same command i need to get one more record that is record number 89. Please let me know how to retrieve that also. I am not looking for all the records from 23 to 89. I am just looking for specific records 23 and 89 only

Regards
Rakesh

Yout can do this with two commands

awk 'NR==23; NR==89'

or with a logical or

awk 'NR==23 || NR==89'

This is the short condition{action} form where the default action is print .

Thank you so much. This is what I am looking for. Thanks alot.

If the input file has a lot more than 89 records, you might also want to try:

awk 'NR==23;NR==89{print;exit}'

to avoid reading the rest of the file after line/record 89.

Thanks alot!