php response in ajax

Question

I have a IP range php function that responds with 1 if ip is in range.

Php:

<?
$ip = $_SERVER['REMOTE_ADDR'];

$public_ip_ranges = array();

$range = (object) array();
$range->name = 'Barclays';
$range->lower = '141.228.0.0';
$range->upper = '141.228.255.255';
$public_ip_ranges[] = $range;

$range = (object) array();
$range->name = 'Incisive Media';
$range->lower = '10.1.0.0';
$range->upper = '10.1.255.255';
$public_ip_ranges[] = $range;

$range = (object) array();
$range->name = 'Barcap';
$range->lower = '146.127.0.0';
$range->upper = '146.127.255.255';
$public_ip_ranges[] = $range;


if (($lngIP=ip2long($ip)) < 0) $lngIP += 4294967296;  

foreach ($public_ip_ranges as $ip_range) {
   if (($lngLow=ip2long($ip_range->lower)) < 0) $lngLow += 4294967296;
   if (($lngHigh=ip2long($ip_range->upper)) < 0) $lngHigh += 4294967296;
   if($lngIP >= $lngLow and $lngIP <= $lngHigh) {
       echo 1;
    }

}

?>

I have ajax referring to this file but what i would like to do is if PHP function echo's 1 then display x text else display Y text.

Ajax:

   $.ajax( { 
            url: "/microsub.php",
            method: 'GET',
            success: function (data) {
             console.log(data);
             },
             error: function(error) {
                 console.log(error);
             }

        } );

So I would like to take the response from the php (which should echo 1 if in range) and display text alert else if not display different text alert

And your question is...? This site is for questions, not a place to dump to-do lists. — Marc B
– Marc B, Commented Aug 9, 2016 at 16:25
You might want to try something like this in your success: function(data) part: if(data == 1){ // do something because the IP is in range } else { // do another thing because the IP is not in range }. It's like Marc says, though, we need your code to help you with. — Henders
– Henders, Commented Aug 9, 2016 at 16:25

buildok · Accepted Answer · 2016-08-09 18:57:06Z

1

you can return an error:

$isPublic = true;
foreach ($public_ip_ranges as $ip_range) {
   if (($lngLow=ip2long($ip_range->lower)) < 0) $lngLow += 4294967296;
   if (($lngHigh=ip2long($ip_range->upper)) < 0) $lngHigh +=4294967296;

   $isPublic = ($lngIP >= $lngLow) && ($lngIP <= $lngHigh) && $isPublic;
}

if(!$isPublic) {
   header("HTTP/1.0 404 Not Found");
}

echo $isPublic;

answered Aug 9, 2016 at 18:57

buildok

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