Need ideas on how to grep for pattern using awk

rahulbgl · 03-04-2007, 02:01 PM

Hi All,
I have a text like this
002000 2010 4302
002020 1234 5689
002040 1234 5689
*
002060 1034 5689
002080 1234 5689
002100 1234 5689
002120 1234 5689
*
002140 1334 5689
.
I am trying to write an awk script where in i need to get the text line next to the line printed by star namely
002000 2010 4302 (First line always)
002060 1034 5689
002140 1334 5689
.
Any ideas?
Thanks,
Rah

ilikejam · 03-04-2007, 02:37 PM

Hi.

This should do it:

Code:

awk 'NR<2 {print}; /^\*$/ {getline; print}'

The 'NR<2 {print}' prints the first line. The '/^\*$/ {getline; print}' finds lines consisting only of a '*', then prints the next line.

Dave

7stud · 03-04-2007, 08:17 PM

Here's my attempt:

awk.scr:

Code:

BEGIN { prevLine="" } #not necessary since awk will automatically initialize to ""

#MAIN
{
        if(prevLine=="*")
                print $0;
        else if(prevLine=="") #prevLine is only blank for 1st line(see BEGIN)
                print $0;

        prevLine=$0;
}

command:

Code:

 awk -f awk.scr data.txt

data.txt:

Quote:

002000 2010 4302
002020 1234 5689
002040 1234 5689
*
002060 1034 5689
002080 1234 5689
002100 1234 5689
002120 1234 5689
*
002140 1334 5689

output:

Quote:

002000 2010 4302
002060 1034 5689
002140 1334 5689

7stud · 03-05-2007, 10:29 AM

...and with sed:

Code:

sed -n -f sed.scr data.txt

The -n option suppresses output, which an be overridden with the p function when necessary.

sed.scr:

Code:

1p            prints first line
/*/{          searches for a line with a * on it and if found executes the commands in the braces  
        n;    outputs pattern space if the -n option isn't set, then replaces pattern space with next line
        p;    prints the pattern space
}

ilikejam · 03-05-2007, 03:07 PM

Ah, the joys of Unix - 80 ways to do everything.