extract certain parts from a file


 
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# 8  
Old 10-17-2011
No problem. Just a small tweek:

Code:
awk  '
    BEGIN { i = 0; }

    {
        split( $2, a, ":" );               # divide field 2 into hr min sec 
        t = (3600 * a[1]) +  (60 * a[2]);  # compute sec past midnight; 80100 is 22:15:00
        if( t < 80100 )                 # wrapped to next day; must roll
            roll = 1;

        if( snarf || t >= 80100 )       # snarfing or past the magic time
        {
            if(  t >= 80100 && roll )   # 22:15 the next day; clear first
            {
                roll = 0;
                delete capture;
                i = 0;
            }

            snarf = 1;
            capture[i++] = $0;     # buffer the record
        }
    }

    END {
        for( j = 0; j < i; j++ )
            print capture[j];
    }'  input-file

Thinking on your other question.
This User Gave Thanks to agama For This Post:
# 9  
Old 10-17-2011
Thanks a lot agama. It works fine.

and i found the solution for the other ques. just one more print command to print it.

gawk 'c-->0;/pattern/{print;c=7}' file1 > file2

---------- Post updated at 07:50 AM ---------- Previous update was at 07:24 AM ----------

the file size is not of concern here as the file size is in few kb and will always be the same.

however the pattern occurs for multiple times and i need the 7 lines following the pattern for each occurance of the pattern.

thanks for the help.
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