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combining awk and sed

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# 1  
Old 02-21-2011
Lightbulb combining awk and sed
Hi experts,
I have a requirement, In which I need to display the first and last line of a zip file where the line starts with "L". I've writen the code like below using sed and awk.
Code:
gunzip -c 20110203.1104.gz | awk '$1 ~ "^L" {print substr($0,178,15)}' | sed -n '1p;$p'

Is it possible to do it awk only ...? or my commands is good one and don't need any changes ..

Waiting for your suggestions.
Thanks
senthil
# 2  
Old 02-21-2011
I feel your sed part alone will work gr8, it is not.

gunzip -c 20110203.1104.gz | sed -n '1p;$p'
# 3  
Old 02-21-2011
No, I need only lines which starts with L, and i need to substring the line from 178 to 15 character,

thanks
# 4  
Old 02-21-2011
Try:
Code:
gunzip -c 20110203.1104.gz | 
awk '/^L/{print substr($0,178,15);f=1} f && /^L/{s=$0} END{print substr(s,178,15)} '

# 5  
Old 02-21-2011
It retruned complete file

Code:
 
$>^L/{print substr($0,178,15);f=1} f && /^L/{s=$0} END{print substr(s,178,15)}' | wc -l <
1022
$>history | tail -2
147 gunzip -c 20110203.1104.gz | awk '/^L/{print substr($0,178,15);f=1} f && /^L/{s=$0} END{print substr(s,178,15)}' | wc -l
148 history | tail -2
$>zgrep -c  "^L" 20110203.1104.gz 
1021

Is it anything worng i'm trying
Last edited by senthil.ak; 02-21-2011 at 08:14 AM.. Reason: code tag missing
# 6  
Old 02-21-2011
Sorry, didn't test the command, try this one:
Code:
gunzip -c 20110203.1104.gz | 
awk 'f && /^L/{s=$0;next} /^L/{print substr($0,178,15);f=1} END{print substr(s,178,15)}'

This User Gave Thanks to Franklin52 For This Post:
senthil.ak
# 7  
Old 02-21-2011
Great it works perfectly.

Code:
 
$>gunzip -c 20110203.1104.gz | awk 'f && /^L/{s=$0;next} /^L/{print substr($0,178,15);f=1} END{print substr(s,178,15)}' 
000000048973373
000000048974393

Could please explain the command for me.
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