使用php和ajax根据第一个字母选择数据库项时出错

Question

I want to create a simple program which function is: The user chooses a letter from a simple form and then the program generates all the elements from my "prestashop" named database which start with this letter. I created the database in phpmyadmin and created the code in php and in ajax. But I'm not sure if this piece of my code really makes the selection of the elements which start with the letter chosen from the user.An error occurs anyway. Here is my php code:

<?php
$q = intval($_GET['q']);/* 54-th row*/

$con = mysqli_connect('localhost','root','','prestashop');
if (!$con) {
    die('Could not connect: ' . mysqli_error($con));
}

mysqli_select_db($con,"prestashop");
$sql="SELECT * FROM prestashop WHERE marca LIKE '".$q."'";
$result = mysqli_query($con,$sql);

echo "<table>
<tr>
<th>Marca</th>
<th>Descrizione</th>
</tr>";
while($row = mysqli_fetch_array($result)) {/* 70-th row*/
    echo "<tr>";
    echo "<td>" . $row['marca'] . "</td>";
    echo "<td>" . $row['descrizione'] . "</td>";
    echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>

Is there something I should change to make this works?

</div>

Answer 1

Well, first of, correct syntax for "like" in mysql is:

LIKE '%foo%'
LIKE '%foo'
LIKE 'foo%'

Secondly, according to PHP documentation regarding mysqli_query:

Returns FALSE on failure. For successful SELECT, SHOW, DESCRIBE or EXPLAIN queries mysqli_query() will return a mysqli_result object. For other successful queries mysqli_query() will return TRUE.

So you will need to check what it returns and if it is false, show some error for example.