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Editorial Team
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Asked: 2026-05-25T10:24:56+00:00

I have two tables: users, images . The users has a column paid which

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I have two tables: users, images. The users has a column paid which is a digit for how many images they paid to enter into the contest..
images has a column entered with a value of 0 for not entered, and 1 for entered. I need the sum of these for each user.

I am trying to query all users who paid for more images than they entered. but it’s not working correctly.. heres what prints:

bob@example.com        8  2  
jcannon@example.com    3  3  
jrcih@example.com      7  3  
aerwqeerll@example.com 5  1  
ray@example.com        2  3

You can see that jcannon@ and ray@ should not be in the list, because paid is not greater than entered

here is the code:

$SQL = mysql_query("SELECT  *, SUM(images.entered) 
                        FROM users, images 
                        WHERE users.id=images.owner 
                        AND users.paid>'SUM(images.entered)' 
                        GROUP BY users.id  ");

while($sendto=mysql_fetch_array($SQL)){
    print $sendto['email']." ".$sendto['paid']. 
          " ".$sendto['SUM(images.entered)']. "<br>";
}

What am I doing wrong?

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1 Answer

  1. Editorial Team

    First get rid of the quotation marks, they are for literal strings. Second, you can only compare aggregated values with the HAVING clause:

    … WHERE … GROUP BY … HAVING users.paid >= SUM(images.entered) …

    Also I find it often a lot clearer to use explicit join syntax, instead of WHERE:

    SELECT  *, SUM(images.entered)
FROM users
INNER JOIN images
ON users.id = images.owner
GROUP BY users.id
HAVING users.paid > SUM(images.entered)
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