awk getline to display first three lines in file
Normand Marion
normand.marion at gmail.com
Wed Dec 30 19:03:21 UTC 2015
awk '{ if ( NR <= 3 ) print $0 }' your_file
2015-12-29 10:12 GMT-05:00 asad <a.alii85 at gmail.com>:
> I'm using
>
> awk 'BEGIN {while((getline gf < "awk.script") > 0) {print gf}}'
>
> to read me first 3 three lines of the pattern. I can do with head -3 . But
> I don't want to since, I have to switch lines between two files.
>
> Suppose awk.script contains
>
>
> line 1
> line 2
> line 3 <------- until here
> line 4
>
> I have tried updating the original cmd with this
>
> awk 'BEGIN {while((getline gf < "awk.script") > 0) && f{print
> $gf;f=0};NR<=3{f=1}}'
>
> the code is influenced from the examples give on link:-
>
> http://stackoverflow.com/questions/17908555/printing-with-sed-or-awk-a-line-following-a-matching-pattern
>
> error i'm getting is
>
> awk: cmd. line:1: BEGIN {while((getline gf < "awk.script") > 0)
> f{print $gf;f=0};NR<=3{f=1}}
> awk: cmd. line:1: ^
> syntax error
> awk: cmd. line:1: BEGIN {while((getline gf < "awk.script") > 0)
> f{print $gf;f=0};NR<=3{f=1}}
> awk: cmd. line:1:
>
> How can it be done? thanks
>
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--
*Normand Marion*
normand.marion at gmail.com
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