mysql_query returning error when using LIKE

Question

I am trying to create a script that selects all users in a given MySQL Column by a given term and city. The city will be specific to the member but each member could have 3 or 4 different positions (bartender, server, host etc). The code I am trying to use is below, however, it is giving me an error. Let me know if you need more information. Thanks!

Could not find staff: Unknown column 'Bartender' in 'where clause'

 

<?php

    $perpage = 15;
    $city = $_GET['city'];
    $type = $_GET['type'];

    if(isset($_GET["page"]))
    {
        $page = intval($_GET["page"]);
    }
    else
    {
        $page = 1;
    }

    $calc = $perpage * $page;
    $start = $calc - $perpage;
    $result = mysql_query("SELECT * FROM staff WHERE titles LIKE $type AND city=$city LIMIT $start, $perpage");
    $rows = mysql_num_rows($result);
    if($rows)
    {
        $i = 0;
        while($post = mysql_fetch_array($result))
        {
?>
            <tr style="background-color: #cccccc;">
                <td style="font-weight: bold;font-family: arial;"><?php echo $post["staffnum"]; ?> >> <?php echo $post["titles"]; ?></td>
            </tr>
            <tr>
                <td style="font-family: arial;padding-left: 20px;"><?php echo $post["abt1"]; ?></td>
            </tr>
<?php
        }
    } else {
          die('Could not find staff: ' . mysql_error());
    }
?>

Answer 1

In order to use LIKE as you want, you need to wrap it in quotes and use the appropriate % characters.

$type = mysql_real_escape_string($_GET['type']);
// do the same with $city and any other user input

$result = mysql_query("SELECT * FROM staff WHERE titles LIKE '%" . $type . "%' AND city='" . $city . "' LIMIT $start, $perpage");