Shell Programming and Scripting

I need to be able to search for a beginning line header, then use grep or something else to get the very next instance of a particular string, which will ALWAYS be in "Line5". What I have is some data that appears like this:

[1] Line1
Line2
Line3
Line4
Line5
Line6
Line7

[2] Line1
Line2
...

In my case, there are up to 1000 numbered beginner lines. I will be keying off of data from line 1, and I know the information I want is in Line5, every time. So, how do I search for the relevant beginning "Line1", then extract the next instance of "Line5" to use?

Thanks!

Do you mean a literal 'Line 5' or a five line offset from the start of a datablock?

If yes,

Code:

grep '^Line5' inputfile

else

Code:

awk ' /^Line1/ { ok=FNR+5}
        /FNR==ok/ {print $0}' inputfile

In the latter case you need to change "Line1" to the actual datum you use to find the start.

Something like this might work for you:

Code:

#!/usr/bin/env ksh

awk -v pat="[$1]" '
    $1 == pat {
        snarf = 1;
        next;
    }

    snarf {
        if( ++snarf == 5 )
        {
            print;
            exit(0);
        }
    }
' $2
exit

The script assumes you pass 1, 2, 3, etc. on the command line which then finds the section "[1]" etc in the input file. The second, optional, command line parameter is the name of the input file. If not supplied, the script will read from stdin. It finds the section and then prints the fifth line that follows.

If you key string isn't of the form [n], then just change the -v pat="" string to suit your needs.

Assuming you put this script into foo.ksh, the following would invoke it to find the 986th section from "input.file"

Code:

foo.ksh 986 input.file

Try:

Code:

awk '/^\[/{for (i=1;i<5;i++)getline;print}' file

Code:

sed -n '/^\[1\]/{n;n;n;n;p}' inputFile

--ahamed