php: change value of variable based on dropdown list

Learning PHP and having an issue that I can’t figure out. I have read that PHP only has scope for functions, so I’m not sure why my switch statement isn’t changing the value of variables.

Goal: to change the mysql SELECT statement based on user selection of drop-down.

Form:

    <form action="contacts_show.php" method="POST">
    <select name="grade" id="grade">
    <option value="all">All Levels</option>
    <option value="elementary">Elementary</option>
    <option value="middle">Middle</option>
    <option value="senior">Senior</option>
<input type="submit" name="browse" id="browse" value="Browse" />
</form>

PHP (edited to shorten code):

$levelSelected = $_POST['grade'];

if ($levelSelected == "all") {
    $querySelect = "SELECT * FROM teachers ORDER BY school ASC";
} else {
    $querySelect = "SELECT * FROM teachers WHERE school LIKE %$levelSelected% ORDER BY school ASC";
}

$query = $querySelect;
$result = mysqli_query($connection, $query);
confirm_query($result);

the confirm_query function, if needed:

    function confirm_query($result_set) {
    if (!$result_set) {
        die("Database query failed.");
    }
}

When “All Levels” from drop-down is selected, code runs as expected. When any other option is selected, my confirm_query function states that the query fails.

I’m not sure why the variable’s values are not switching.

Answer

To elaborate on my comment:

Change LIKE %elementary% to => LIKE '%elementary%' and do the same for the others.

You need to wrap the pattern match in quotes, and as per the manual:

• http://dev.mysql.com/doc/refman/5.0/en/string-comparison-functions.html#operator_like

mysql> SELECT ‘David!’ LIKE ‘%D%v%’;
mysql> SELECT 10 LIKE ‘1%’;

You’re also not checking for DB errors.

Add or die(mysqli_error($connection)) to mysqli_query()

If that still doesn’t work, then it’s also a scope issue.
Pass the connection to your function, do not make it global.