Need to create shell script that finds a phrase and then prints the line right above that phrase to another file

I need to create a shell script (or perl) that will search a given logfile for a particular phrase and then copy the date and time of which is just above the occurance of the phrase.

Example:

11/26/2003;10:31:05
Verification Startup is completed
Welcome to the Dialogue
11/26/2003;10:32:11 Transition started
11/26/2003;10:32:14 Manipulation error shutdown completed
11/26/2003;10:41:25
Verification Startup is completed
Welcome to the Dialogue

Given the logfile above I need to write another file that will contain this:

Startup times
11/26/2003;10:31:05
11/26/2003;10:41:25

Please help, thanks. I am writing this in an HP9000 machine. Unix.

jporgyAsked:
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jporgyAuthor Commented:
This is a unix perl or shell script programming
0
sunnycoderCommented:
echo Startup times > output_file_name
grep "string_to_search" log_file_name | awk ' { print $1 " " $2 } ' >> output_file_name
0
shivsaCommented:
#!/usr/bin/perl

require Exporter;

my ( $line_no,
     $one_line,
     $true,
     );

$line_no = 0;
print "Verification Startup \n";
@aa = ();
@aa = `cat /tmp/historydmp.txt`;
while ($line_no < @aa ){
$one_line = @aa[$line_no];
$true = grep /Verification/, $one_line ;
if ( $true == 1) {
print "@aa[$line_no-1] \n";
     }
$line_no = $line_no + 1;
}
0

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jporgyAuthor Commented:
shivsa's answer is okay.
0
jporgyAuthor Commented:
My html screen is screwed, can the moderator award the answer to shivsa'?
0
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