PHP mysql查询计算左连接中的重复值

Question

I have two tables - projects and sites.

I want to show a table with project number and site addresses, but where there's multiple sites per project show "Multiple Sites", and where there's no site assigned to a project show "No site".

So I would like my table to look like this:

Project Number	Site Address
job1	123 Fake Street
job2	5 sydney street
job3	Multiple Sites
job4	No site

But currently my table looks like this.

Project Number	Site Address
job1	123 Fake Street
job2	5 sydney street
job3	1 First Street
job3	2 Second street
job4	No site

with job3 repeating. My code looks like this:

<?php
    $query = mysqli_query($conn, 
        "SELECT projects.project_number,
            sites.address,
            sites.project_fk
            FROM projects
            LEFT JOIN sites
            ON projects.id = sites.project_fk");
?>

<div class="container">
    <table class="table">
         <thead>
            <tr>
            <th scope="col">Project Number</th>
            <th scope="col">Site</th>
            </tr>
        </thead>
        <tbody>
          <?php
            while ($row = mysqli_fetch_array($query)) {
                echo "<td>".$row["project_number"]."</td>";
                if ($row["address"] == null) {
                echo "<td>No Site</td>";
                } else {
                echo "<td>".$row[1]."</td>";
                }
                echo "</tr>";
                }
            ?>
        </tbody> 
    </table>
</div>

Is there a way to do this? Do I have to create a new query which counts the number of 'sites' which are the same, and if so how do I do this?

Thanks!

Answer 1

Use group by and case to determine if there are multiple, single or none addresses

SELECT projects.project_number,
        CASE
        WHEN COUNT(*) > 1 then 'Multiple sites'
        WHEN COUNT(*) = 1 then sites.address
        WHEN COUNT(*) = 0 then 'No site'
        END as address
FROM projects
LEFT JOIN sites ON projects.id = sites.project_fk
GROUP BY projects.id