Array not inputting data from database

Question

I'm trying to make an array with the structure like so:

[ 
    { 
        "ips": 
             "xxx.xxx.xxx.xx", 
             "yyy.yyy.yyy.yy",
             "zzz.zzz.zzz.zz" 
        "ai": 
             "xxx.xxx.xxx.xx",
             "yyy.yyy.yyy.yy"
    } 
]

So I can check if variable with string is inside array. But I'm collecting the IP's from a DB using mysql, and am not sure how to insert the IP's.

I've tried this:

$whitelist = array(
    'ips' => array(), 
    'ai' => array()
);

// sql command
$sql = "SELECT `ip` FROM `testers`";
$res = mysql_query($sql) or die(mysql_error());

while($row = mysql_fetch_assoc($res)) {
    array_push($whitelist['ips'], 
        $row
    );
}

Can somebody please tell me how I can do this/ Seriously so lost other then inserting the IP's individually. That would take forever and not easy to track.

That's not a valid PHP array

John Conde
– John Conde

2014-04-12 02:18:45 +00:00
Commented Apr 12, 2014 at 2:18 — John Conde
– John Conde, Commented Apr 12, 2014 at 2:18

Brett · Accepted Answer · 2014-04-12 02:48:16Z

1

It looks like your solution is really close. You just need to access the row that you retrieved from mysql as an associative array using $row['ip'].

The below code should save the ip's from the database to the $whitelist['ips'] array and then print the results to the screen.

$whitelist = array(
    'ips' =>array()
);
$sql = "SELECT `ip` FROM `testers`";
$res = mysql_query($sql) or die(mysql_error());
while($row = mysql_fetch_assoc($res)) {
    array_push($whitelist['ips'], $row['ip']);
}
var_export($whitelist);

edited Apr 12, 2014 at 2:48

answered Apr 12, 2014 at 2:28

Brett

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