[SOLVED] Need help with comparing LIKE results

[pneudralics]

Having issues with my if and else statment. I'm not sure if I'm comparing them correctly. When I submit a name I want to check the database if the database has similiar names. If there are less names than zero I want to echo zero. Am I writing it correctly?

 

Thanks

 

$name = strip_tags(htmlentities(trim($_POST['name'])));
$namesq = "SELECT * FROM names WHERE name LIKE '%$name%'";

if ($namesr = mysql_query ($namesq)) {
while ($namesrow = mysql_fetch_array ($namesr)) {
$name = $namesrow['name'];
}
}

//if zero result echo zero
if ($name < 0) {
echo "zero";
}
//else if more than zero result echo found
else {
echo "found";
}

[hellonoko]

I think you would want to use mysql_num_rows() http://au.php.net/mysql_num_rows

 

To count the number of results then compare that to 0-?

[JonnoTheDev]

I have tidied this up for you

<?php
if(!$result = mysql_query("SELECT name FROM names WHERE name LIKE '%".mysql_real_escape_string(strip_tags(trim($_POST['name'])))."%'")) {
die(mysql_error());
}
$names = array();
while($row = mysql_fetch_array($result)) {
// add each name to the $names array
$names[] = $row['name'];
}
echo (count($names) ? "Results Found" : "No Results Found") ;
?>

[Potatis]

Your code is beautiful, Neil. It's like php poetry.