Jump to content


This topic is now archived and is closed to further replies.


LIKE query question

Recommended Posts


I am getting a 0 when echoing the variable from the code snippet below, and I am sure it has something to do with the LIKE portion on the SQL query, but I cannot figure it out.. The goal is to print the percentage of free IPs on our local network, but it's showing a zero. Is the answer glaringly simple?
    $ipb10 = '10.';
    $query1 = "SELECT SUM(no_ips) AS totalip FROM ips WHERE INET_NTOA(sip) LIKE '.$ipb10.' ";
    $result = mysql_query($query1);
    $r = mysql_fetch_array($result);
    $free = 16777216;
    $pctg1 = ($r['totalip']/$free*100);
    echo number_format($pctg1, 2);[/code]

Share this post

Link to post
Share on other sites
I tried that and am still getting a zero =/


My own fat finger fault, that works

/bonk self

Thank you kindly!

Share this post

Link to post
Share on other sites


Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.