getenv() function question

why doesn't this work?

[madcat]

In my book, it has a couple of examples to follow along with- it is suggested to use:

<?php 
 $address = getenv("REMOTE_ADDR"); 
 echo "Your IP address is $address."; 
?>

But only this will work:

<?php 
 $address = $_SERVER['REMOTE_ADDR']; 
 echo "Your IP address is $address."; 
?>

--> wrapping getenv() around the latter fails too.

So, should I be using getenv()?
Why doesn't it produce the users IP address in this case?

php.net mentions this form:

 if (getenv(HTTP_X_FORWARDED_FOR)){  
 $ip=getenv(HTTP_X_FORWARDED_FOR);  
 }  
  else {  
  $ip=getenv(REMOTE_ADDR);  
  }

Thanks for any clarity-

[SmallTime]

I think it depends whether register_globals is on or off, default used to be on, now off. Check php.info

[madcat]

I'll check it out really...but php.net has as example:

<?php
$address = getenv('REMOTE_ADDR');
echo "Your IP address is $address.";
?>

>> Your IP address is .
-------------------------------------------------
Using:

<?php
$address = $_SERVER['HTTP_USER_AGENT'];
echo "Your IP address is $address.";
?>
OR...
<?php
$address = $_SERVER['REMOTE_ADDR'];
echo "Your IP address is $address.";
?>

Work correctly- so are they wrong to use? Is getenv() necessary? I'm using PHP 4.2.2- just a bit confusing when your trying to learn.

Thanks-

[dhdweb]

Why not just do this:

<?
$address = "$REMOTE_ADDR";
echo "Your IP is $address";
?>

[toadhall]

dhdweb

If register_globals is turned off $REMOTE_ADDR won't work. You have to use $_SERVER['REMOTE_ADDR'];

[madcat]

I did the latter- I guess I don't need to use getenv() to extract the value and place it in a variable-- it's just what php.net and my author suggest- so i thought it was the right thing.