I'm trying to create a simple script that logs access to several pages
on a site I'm working on. I want to capture the date and time, address
of the PC accessing the page, and the name of the script being accessed.
I have the following code:
<?php
$today = date("F j, Y H:i:s");
$lineOut = $REMOTE_ADDR;
$lineOut .= " ";
$lineOut .= $today;
$lineOut .= " ";
$lineOut .= $SCRIPT_NAME;
$lineOut .= "\n";
$fp = fopen("logfile.txt","a");
fputs($fp,$lineOut);
fclose($fp);
?>
Everything works fine except for the line "lineOut .= $SCRIPT_NAME;"
The value of $SCRIPT_NAME is always the path to the PHP executable, not
the script that contains the code. How do I correct this problem. I'm
running PHP 4.06 under Apache 1.3.22 on a Win 2000 PC.
Thanks in advance,
Brad
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