I got it figured out. I needed $PHP_SELF instead.
Brad
Brad Harriger wrote:
> 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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