MySQL query malfunctioning?

I have the below MySQL query, which should return the t1.username,, in_count (WHERE t2.type = 'in'), out_count (WHERE t2.type = 'out') y WHERE esta campo es NOT EMPTY or NULL.

However its seemed to have malfunctioned (or I'm doing something wrong?), as it returns 1 result (via mysql_num_rows()) and when i print_r() (sobre el mysql_fetch_assoc()), the keys of the columns are their however all the values are empty/blank (see below).

La print_r() dump:

    [username] => 
    [website] => 
    [in_count] => 
    [out_count] => 

La consulta de MySQL:

SELECT t1.username,,
       SUM(IF(t2.type = 'in', 1, 0))  AS in_count,
       SUM(IF(t2.type = 'out', 1, 0)) AS out_count
FROM   users AS t1
       JOIN referrals AS t2
         ON t1.username =
WHERE  NOT Isnull(
LIMIT  0, 10 

Hope all is clear, let me know if you need more information, appreciate all help. :B

preguntado el 09 de enero de 11 a las 06:01

2 Respuestas

Sounds like you might need a GROUP BY clause at the end? Perhaps

GROUP BY t1.username,

That way you'll get all the in and out counts for a particular user on a particular web site.

I was under the impression that with aggregate functions like SUM the query requires a GROUP BY clause... I'm surprised the server accepted your SQL.

Respondido el 09 de enero de 11 a las 09:01

By design, sadly; SQLite is the only other DB that supports the functionality. - Ponis dios mio

La SUM() function is an aggregate, and there is no GROUP statement in your query, so it has nothing to do. You can always check the error condition of your query in PHP by outputting the contents of mysql_error(), which in this case will likely say that you are missing a GROUP declaración.

Based on your query it looks like your intent is to have a one-to-many relationship between users and referrals, so you should be grouping by something like t1.username (or a user ID if your table has it).

Respondido el 09 de enero de 11 a las 09:01

No es la respuesta que estás buscando? Examinar otras preguntas etiquetadas or haz tu propia pregunta.