CODEWITHSUNDEEP
linuxawk
Anuj vijjan
Anuj vijjan

Reputation: 11

Linux awk command to retrieve first and last row

I'm trying to retrieve the first and last row of awk command, but I'm not able to. Where I'm making mistakes.

egrep 'updateAll|update-mgr|Startup REX|configd.*UnitProperty updated' $LOG | \
  awk 'NR == 1{print $1" "$2" "$3} NR == $#{print $1" "$2" "$3}'

awk: cmd. line:2: NR == 1{print $1" "$2" "$3} NR == $#{print $1" "$2" "$3}
awk: cmd. line:2:                                    ^ syntax error

I expect first row and last row form the output of grep command.

Upvotes: 1

Views: 2179

Answers (3)

tink
tink

Reputation: 15206

Btw - most of the time when you use grep & awk in one command you're doing it wrong ;)

awk '
{  
    if ($0~/updateAll|update-mgr|Startup REX|configd.*UnitProperty updated/) {
       c++
       l=$0
       if(c==1){print}       
    }
}
END{print l}
'  $LOG

Upvotes: 2

Dudi Boy
Dudi Boy

Reputation: 4865

@tink solution is correct. I just wish to add that: .

Using variable l is bad practice, since it can be mistaken to 1 or I depending on the current font.

The default match pattern can be extracted as filter pattern for record processing.

So @tink solution becomes:

awk '
/updateAll|update-mgr|Startup REX|configd.*UnitProperty updated/ {  
    c++;
    lastLine = $0;
    if(c == 1){print}       
}
END{print lastLine}
'  $LOG

Upvotes: 0

afenster
afenster

Reputation: 3608

Use NR == 1 to get the first line, and END to get the last line.

Having this input:

$ echo -e '1 a v\n2 b w\n3 c x\n4 d y\n5 e z'
1 a v
2 b w
3 c x
4 d y
5 e z

we can parse it using this awk command:

$ echo -e '1 a v\n2 b w\n3 c x\n4 d y\n5 e z' | awk 'NR == 1 { print $2; } END { print $2; }'
a
e

Upvotes: 0

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