将返回的SQL数组与随机结果选择器一起使用

Question

I am writing a new feature plugin for a website I develop and am stuck with how to go about implementing such feature. I'm trying to return an SQL query to an array that I can then pass to a function in either PHP or JavaScript, to perform a random option picker, and display the results to the user. I'm trying to perform the random picker by an on-click functionality.

So far, I have figured out how to successfully select my SQL query, return statements, and in JavaScript, work with a random option picker. However, I have not been able to find a good way to combine all these things together to produce my result.

I was wondering if anyone knew of a way to do this. Is it better to perform the entire program functionality in PHP or better to pass it to a JavaScript function and return results through the JS code?

Here are some examples of my code. Please assume that my query works. I do not need the display option in my code. It is implemented to check query results. I was hoping to hide the array in the real plugin from users and use the returned results for the on-click functionality attached to an HTML element.

<?php
    function choices() {
        $config = parse_ini_file(*database file*);
        $con = mysqli_connect(*database information*);
        if(!$con){
            exit("Please try again later."); 
        }

        $query = "SELECT * FROM *query*";
        $result = @mysqli_query($con, $query);
        $num = mysqli_num_rows($result);
        if ($num > 0) {

            echo "<p>There are currently $num restaurants.</p>\n";

            echo '<table width="60%">
            <thead>
            <tr>
                <th align="left">Restaurant Title: </th>
                <th align="left">Link: </th>
            </tr>
            </thead>
            <tbody>
        ';

            while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
                echo '<tr><td align="left">' . $row['post_title'] . '</td><td align="left">' . $row['guid'] . '</td></tr>
                ';
            }

            echo '</tbody></table>';

            mysqli_free_result($result);
        }

        else {
            echo '<p class="error">The results could not be retrieved.</p>';
        }

        mysqli_close($con);
    }

    $foo = choices();
?>

My JS code.

var favorites = ["choice #1, choice #2, choice #3, choice #4"];
var favorite = favorites[Math.floor(Math.random() * favorites.length)];

Thank you very much for any input regarding this question! I understand this is a rather lengthy question and I do appreciate time took to help me understand it.

Answer 1

Why not just have your database return a single random result since that's all you need?

SELECT * FROM tbl ORDER BY RAND() LIMIT 1

Take a look at this post about getting random rows

Answer 2

Put the results of the query in a PHP array. Then use json_encode() to convert that to a Javascript array. Then the onclick functionality can display a random element of the array.

$array = array();
while ($row = mysqli_fetch_assoc($result)) {
    $array[] = $row['post_title'];
}
?>
<script>
var favorites = <?php echo json_encode($array); ?>;
$("#button").click(function() {
    var favorite = favorites[Math.floor(Math.random() * favorites.length)];
    alert(favorite);
});
</script>