[SOLVED] Having a problem with a Query

[JSHINER]

<?php

session_start();

require('php/functions.php');

require('php/Database.php');

$page = array();



function Get_Results($db, $n = false, $c = false) {

$arr = $db->getArray("SELECT * FROM table WHERE name LIKE %$n% AND company LIKE %$c% AND category_id = 81");

return $arr;

}

$db = new Database();

$page['results'] = Get_Results($db, $_GET['n'], $_GET['c']);


$db->close();


// Results


echo 'Search Results';

foreach ($page['results'] as $r) {

echo'Name:', $r['name'], ' - Company: ', $r['company'], ''; }


//--------

?>

 

I am using a form to submit to the above page with "n" and "c" submitted in the form. To test I simply put "page.php?n=John&c=Company" in the address bar, and get the following error:

 

Notice: Error in database query. Query was [sELECT * FROM table WHERE name LIKE %John% AND company LIKE %Company% AND category_id = 81] and the error returned was [You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '%Sandra% AND company LIKE %Keller% AND category_id = 81' at line 1] in folder on line 71

Search Results

Warning: Invalid argument supplied for foreach() in folder/page.php on line 34

 

 

Does anyone know where my errors are here?

[GingerRobot]

Try:

 

$arr = $db->getArray("SELECT * FROM table WHERE name LIKE '%$n%' AND company LIKE '%$c%' AND category_id = 81");

 

If you are searching for a string, you have to enclose the string in quotes.

[JSHINER]

I still get the same error, it just now puts ' ' around the '%JOHN%'

[JSHINER]

Problem resolved. Needed to do '%$n%'