google refferer

[chrismarsden]

Hi guys, can you help me...

 

im using the below code and i have been trying to figure out how to make it so that if a known ip searchs for my site it gives an output of "from the office" instead of the IP... any ideas?

 

 <?php

$name="Chris"; 
$email_address="email address here";
$email_address2="email address here";
$keywords="keywords here";
$referrer=$_SERVER['HTTP_REFERER'];

$ip=$_SERVER['REMOTE_ADDR'];



if( (stristr($referrer, "google")) && (stristr($referrer, "search")) ) {

   parse_str($referrer, $output);
   $keywords=$output['q'];

   $email_message="A visitor just arrived to the website after searching for '$keywords', Their IP address is '$ip'. ($referrer)";

   mail ("$email_address","Google referred a visitor","$email_message");
     mail ("$email_address2","Google referred a visitor","$email_message");
}
?>

 

[chrismarsden]

i know it will be an if statement but not sure how to write it... any ideas anyone?

[Adam]

I'd use in_array to lookup the IP address within an array of known IPs. You could also extend it to store the 'from' text and have multiple known IPs:

 

$known_ips = array(
    '123...' => 'from the office',
    // etc.
);

$ip = $_SERVER['REMOTE_ADDR'];

if (in_array($ip, $known_ips))
{
    $ip = $known_ips[$ip];
    // you may want to use a different var name here now?
}

 

You could even pull the list in dynamically from another data source (e.g. a database).

[chrismarsden]

so the code would look like:

 

  <?php

$name="Chris"; // add your name here
$email_address=""; 
$email_address2=""; 
$keywords="";
$referrer=$_SERVER['HTTP_REFERER'];

$known_ips = array(
    'IP here' => 'from the office',
);

$ip = $_SERVER['REMOTE_ADDR'];

if (in_array($ips, $known_ips))
{
    $ip = $known_ips[$ips];

}


if( (stristr($referrer, "google")) && (stristr($referrer, "search")) ) {

   parse_str($referrer, $output);
   $keywords=$output['q'];

   $email_message="A visitor just arrived to the website after searching for '$keywords', Their IP address is '$ip'. ($referrer)";

   mail ("$email_address","Google referred a visitor","$email_message");
     mail ("$email_address2","Google referred a visitor","$email_message");
}
?>

[chrismarsden]

just realised this does not work... im a noob... can you help me out lol

[Adam]

Can you be more specific with how it doesn't work?

[chrismarsden]

it does not output the line "from the office" in the email, just the IP as it always has done.

[Adam]

Ahh, of course. Sorry my bad. I started using in_array() but when I thought about extending it to store the from text I modified the array to store the IP as the key, but I should have also switched to using array_key_exists:

 

$known_ips = array(
    '123...' => 'from the office',
    // etc.
);

$ip = $_SERVER['REMOTE_ADDR'];

if (array_key_exists($ip, $known_ips))
{
    $ip = $known_ips[$ip];
    // you may want to use a different var name here now?
}

 

That should work for you.

[chrismarsden]

that seems to work brill...