Posted by Paul Lautman on 09/01/07 11:40

Christian Aastorp wrote:
> I'm a newbie with php, but experienced with other languages. I'm
> maintaining a small site written in php. To ease debugging and updates
> I've used an onscreen reference to indicate identity of included files
> etc.
>
> I don't really want this reference to be visible to anyone but myself.
> So I searched the manual and found some promising predefined variables
> that I used in this code snippet:
>
> <?php
> if ($_SERVER['REMOTE_ADDR'] = "127.0.0.1") {
> echo $level1, '_', $level2, '_', $imageNO, ' ', $display;
> }
>>
>
> This worked spllendidly when I tested running our site locally using
> the XAMPP-bundle. But when I copied the code out to the real server it
> seems that all clients are found at 127.0.0.1!
>
> I'm sure there are lots of better ways both to fix my code, and
> especially totally different ways to do what I'm trying to achive, any
> suggestions very welcome!

As Michael pointed out, what you are actuall y doing here is assiging the
value "127.0.0.1" to the associative array element $_SERVER['REMOTE_ADDR'].

You should write:
<?php
if ($_SERVER['REMOTE_ADDR'] == '127.0.0.1') {
echo $level1, '_', $level2, '_', $imageNO, ' ', $display;
}
>
Note the == instead of = and also note the use of single instead of double
quotes around the string, which stops PHP from parsing the string to
discover if any substitution needs to take place.

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