根据PHP中的结果运行第二个MySQL查询

Question

EDIT: I've just thought I could run this as one query with something like this:

SELECT * FROM clubs
JOIN match_request_archive
ON (SELECT COUNT (clubs.id) AS requests FROM match_request_archive)

I appreciate though my syntax is horrendous I've just whipped this up and will tweak it now.

---

I have a basic script for displaying data in a table. I've got a problem I can't solve which is I need to run one simple query to show all the data to the user. But in the last column I need to run a separate query to get the result.

Basically the first column in my table has the unique ID of a Rugby Club on the database. In a separate table on the database is a list of all requests made by clubs to players. I want to display how many requests each club has made in the last column of my script below which means running this query:

SELECT * FROM match_request_archive WHERE club_id LIKE "$row['id']"

The $row['id'] is from the PHP script below:

<?php
include 'credentials.php';

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

$sql = "SELECT * FROM clubs";

$result = $conn->query($sql);

if ($result->num_rows > 0) {

    while($row = $result->fetch_assoc())

    {

        echo "
        <tr>
        <td>" . $row['id'] . "</td>
        <td>" . $row['club_name'] . "</td>
        <td>" . $row['club_captain'] . "</td>
        <td>" . $row['club_captain_number'] . "</td>
        <td>" . $row['club_captain_email'] . "</td>
        <td>TBC</td>
        </tr>
        ";
    }
} else {
    echo "<tr><td>0 results</td></tr>";
}
$conn->close();
?>

To solve this I tried embedding a second script in the echo command but it's my understanding you can't do this and in any case it did cause a fatal error. I can run this query separately and have it displayed in a different table, my problem is I need this to be all in one table. I hope that all makes sense.

Answer 1

Executing query in loop is bad practice because it has very slow performance. So you should try get all data by one query if it possible:

SELECT *, COUNT(*) FROM clubs
LEFT JOIN match_request_archive
ON match_request_archive.club_id = clubs.id
GROUP BY clubs.id