Hi
basic my query is how to print all the output in $msg variable
it should print values of $previousLine, $var as buffered variable
Regards
Sreedhar
-----Original Message-----
From: Kalkunte-Venkatachala, Sreedhar
Sent: 23 October 2004 12:32
To: 'Zeus Odin'
Subject: RE: How to store the out put in StringBuffer
Hi
Basically I have written a program similar to tail -f. System will search for error
message. If found it has to through to chat channel. Problem is it connects to chat
channel to print each line. I want to store in a variable and print at one shot. How
to do this?
Kindly let me know
==========================
open (LOGFILE,"$FILENAME");
for (;;)
{
for ($cursorposition = tell(LOGFILE); $_ = <LOGFILE>; $cursorposition = tell(LOGFILE))
{
$line = $_;
chomp $line;
push @buffer, $line;
if (@buffer > 10)
{
shift @buffer;
}
if ($line =~/$regex/)
{
$j = 0;
foreach $previousLine (@buffer)
{
$j++;
$msg=$previousLine;
}
$i = 0;
$pos = tell LOGFILE;
while(<LOGFILE>)
{
$i++;
if ($i > 10)
{
last;
}
$var = $_;
$msg=$var;
}
DoChat($msg);
seek(LOGFILE, $pos, 0);
}
}
sleep(1);
seek(LOGFILE, $cursorposition, 0);
}
====================================================================
-----Original Message-----
From: Zeus Odin [mailto:[EMAIL PROTECTED]
Sent: 23 October 2004 12:29
To: Kalkunte-Venkatachala, Sreedhar
Subject: Re: How to store the out put in StringBuffer
First, always
use warnings;
use strict;
at the top of each program you write. Second, what are you trying to do
exactly? Third, some observations:
1) You use $j and $pos exactly one time. Why are they present?
2) You seem to be looping to 10 using $i. Why not instead use
for my $i(1 .. 10) { code here }
3) What are you trying to do with <LOGFILE>?
It seems to me that you are printing each item in @buffer, then printing
the
next 10 lines of LOGFILE. Is this what you are trying to do? Are @buffer
and
LOGFILE related at all?
Thanks,
ZO
"Sreedhar Kalkunte-Venkatachala"
<[EMAIL PROTECTED]>
wrote...
Hi
Say I have program like this
If I Say print It prints each line. How to store the both output in one
string buffer ($previousLine and $var)
For example I have written this program
========================================
foreach $previousLine (@buffer)
{
$j++;
print "$previousLine";
}
$i = 0;
$pos = tell LOGFILE;
while(<LOGFILE>)
{
$i++;
if ($i > 10)
{
last;
}
$var = $_;
print $var;
}
========================================
Regards
Sreedhar
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