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getting Notice: Undefined variable: ip how do I do this correctly?

Posted on 2008-10-01
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Last Modified: 2013-12-12
I have a probably simple issue..
I have all errors on and want them on. but I am getting an error message:
Notice: Undefined variable: ip in C:\Program Files\Zend\Apache2\htdocs\babyearth_testing\testip.php on line 65

Not sure why??
<?php
$ip = (isset($_SERVER["HTTP_CLIENT_IP"])) ? $_SERVER["HTTP_CLIENT_IP"] : $_SERVER["REMOTE_ADDR"];
 
echo "<br> Your IP recorded is: $ip";
?>

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Question by:jbrashear72
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4 Comments
 
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Accepted Solution

by:
hielo earned 500 total points
ID: 22616336
try:
<?php
$ip = (isset($_SERVER["HTTP_CLIENT_IP"]) ? $_SERVER["HTTP_CLIENT_IP"] : (isset($_SERVER["REMOTE_ADDR"]) ? $_SERVER["REMOTE_ADDR"] : "EMPTY") );
 
echo "<br> Your IP recorded is: $ip";
?>

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Author Comment

by:jbrashear72
ID: 22616600
That did it. Can you explain what corrected it?
I need to understand so if I load a variable I don't get the error.
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LVL 82

Expert Comment

by:hielo
ID: 22616733
you have:
$ip = (isset($_SERVER["HTTP_CLIENT_IP"])) ?

but the value for $ip should be the value of the whole expression on the right. You are closing the first parenthesis immediately before the question mark, but you really need to close it at the very end of the statement.
$ip = (isset($_SERVER["HTTP_CLIENT_IP"]) ? ... : ... );

furthermore, you also need to make sure that
$_SERVER["REMOTE_ADDR"]

isset.
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