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Posted by Aaron Saray on 09/01/07 14:49
On Sep 1, 6:40 am, "Paul Lautman" <paul.laut...@btinternet.com> wrote:
> 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.
I usually run an if against $_SERVER['SERVER_NAME'] ... so on
localhost the servername is either 'localhost' or like
'mydeveloment.local' - where as the public one will be yoursite.com
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