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Home/ Questions/Q 6245489
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Editorial Team
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Editorial Team
Asked: May 24, 20262026-05-24T12:30:56+00:00 2026-05-24T12:30:56+00:00

Let’s say you have a database with two tables named clients and referrals. TABLE

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Let’s say you have a database with two tables named “clients” and “referrals”.

TABLE clients has two columns: “id” and “name”.
TABLE referrals also has two columns: “id” and “referred_by”

Both “id” columns are PRIMARY_KEY, AUTO_INCREMENT, NOT_NULL

TABLE clients has three rows:

1 | Jack  
2 | Frank  
3 | Hank  

TABLE referrals also has three rows:

1 | 0  
2 | 1  
3 | 2  

Meaning, Jack is client 1 and was referred by no one; Frank is client 2 and was referred by Jack; Hank is client 3 referred by Frank.

The SELECT command I used to display the meaning above was:

mysql_query("SELECT clients.id, clients.name, referrals.referred_by FROM clients INNER JOIN referrals ON clients.id=referrals.id");

while ($row = mysql_fetch_array($result))
{
    echo $row['id'] . " " . $row['name'] . " " . $row['referred_by'] . "<br>";
}

Which outputs:

1 Jack 0  
2 Frank 1  
3 Hank 2  

Now the real question is:

How should I modify the code so that it outputs the name of the referrer instead of their id?

Meaning, it should look like this:

1 Jack  
2 Frank Jack  
3 Hank Frank

Thanks in advance ~

Edit: Make sure to mention how I should update the array as I’m lost on how I should update that whole echo line.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-24T12:30:57+00:00Added an answer on May 24, 2026 at 12:30 pm

    You’re almost there – you just need to join back to the clients table a 2nd time to get the referrer name:

    mysql_query("SELECT clients.id, clients.name, rclients.name as referred_by
    FROM clients 
    INNER JOIN referrals ON clients.id=referrals.id
    LEFT JOIN clients as rclients ON referrals.referred_by = rclients.id");
    
    while ($row = mysql_fetch_array($result))
    {
        echo $row['id'] . " " . $row['name'] . " " . $row['referred_by'] . "<br>";
    }
    
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