search pattern and get line 46 before its occurance

[schocku]

Hi all.
I have a fixed format log file with content as follows :
========
06/18/2003 12:00:00:008 blah blah
06/18/2003 12:00:02:010 blah blah
..
..
..
06/18/2003 12:00:00:016 blah blah
06/18/2003 12:00:03:020 submission 124304
..
..
06/18/2003 12:00:04:200 blah blah
06/18/2003 12:00:05:020 submission 124304
=============

Can you please help me with a script which searches for string "124304" in the file and extracts the 46th line before EACH occurance of the string in the file.

Thanks for your help in advance.

 

[CaKiwi]

Try this

/124304/ {
print a[ix]
}
{
a[ix] = $0
ix++
if (ix>n) ix=1
}

CaKiwi

"I love mankind, it's people I can't stand" - Linus Van Pelt

 

[schocku]

CaKiwi thanks for your response. The script worked perfectly. An additional question. Instead of printing the 46th line before the pattern, can I search for the word "Case:" before the pattern 124304 and print that line.

So if the file is as below
========
06/18/2003 12:00:00:008 blah blah
06/18/2003 12:00:02:010 Case: 111
..
..
..
06/18/2003 12:00:02:016 blah blah
06/18/2003 12:00:03:020 submission 124304
..
06/18/2003 12:00:03:190 Case: 222
..
..
06/18/2003 12:00:04:200 blah blah
06/18/2003 12:00:05:020 submission 124304
=============
the output should be

06/18/2003 12:00:02:010 Case: 111
06/18/2003 12:00:03:190 Case: 222

Thanks again for your help.

 

[CaKiwi]

/Case:/{a=$0}
/124304/{print a}

CaKiwi

"I love mankind, it's people I can't stand" - Linus Van Pelt

 

[schocku]

CaKiwi,
My problem is solved. Thanks for your prompt response.