PHP:Variable in function not working

Question

I'm trying to get dns records of a domain through dns_get_record() but the function doesn't seem to work when a variable is inserted..here's my code

<form action="" method="post">
     <input type="text" name="host" placeholder="Enter IP or Domain"/>
     <select name="dns">
          <option value="DNS_A" selected="selected">A</option>
               ....
          <option value="DNS_ANY">ANY</option>
     </select>
</form>
<?php
$host=$_POST['host'];
$dns=$_POST['dns'];
$type=end(explode('_',$dns));
if ($host==""){
     exit();
}
echo "Results for $host $type record<br />";
$result = dns_get_record($host, $dns);
echo "Result = ";
print_r($result);
?>

but if i put

$result = dns_get_record($host, DNS_A);

instead of

$result = dns_get_record($host, $dns);

it works..help!

@ppeterka That is a constant for once, not a string! Don't quote it. — deceze
– deceze ♦, Commented Dec 1, 2012 at 11:58
@deceze I didn't even think of that - I blindly (and dumbly) defaulted to the usual "not escaped string" issue I saw too many times... — ppeterka
– ppeterka, Commented Dec 1, 2012 at 12:02

dev-null-dweller · Accepted Answer · 2012-12-01 11:55:53Z

3

DNS_A is a constant of value 1, while you are passing string "DNS_A", try by passing value of constant, using $result = dns_get_record($host, constant($dns));

answered Dec 1, 2012 at 11:55

dev-null-dweller

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1 Comment

nitish Over a year ago

it worked..but $host works even without passing a value of constant?

Sébastien Renauld · Accepted Answer · 2012-12-01 11:54:43Z

3

This is because your POST returns a literal string: "DNS_A", if you prefer. This is NOT the same as DNS_A, which is a constant and most likely contains an integer.

You'll need a mapping table for this one. Or just pass the integer value straight off.

answered Dec 1, 2012 at 11:54

Sébastien Renauld

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