[PHP] Re: MySQL: AVG in SELECT From: _lallous (elias_bachaalany <email protected>)
Date: 10/11/01

first of all, when you do AVG(q18!='6') it will make the AVG of either
TRUE=1, FALSE=0
maybe if you select q18, and AVG(q18) and when using the result query you
decide from your PHP code whether to refer to the average or to its value...
example:

$result = mysql_query(" SELECT AVG(q1), AVG(q2), AVG(q3), AVG(q4), AVG(q5),
AVG(q6), AVG(q7),
 AVG(q8), AVG(q11), AVG(q16), AVG(q18) as avg_q18,q18 FROM hr_form2 WHERE
Answer != 0");

<?
while ($row=mysql_fetch_Array($result))
{
  if ($row['q18']==6)
    $value = $row['avg_q18'];
 else
   $value = $row['q18'];
}
?>

hope it helps...

"Tom Churm" <thomas.churm <email protected>> wrote in message
news:3BC5619E.8806F6F1 <email protected>
> hi,
>
> i have a select statement to give me the averages for various mysql
> columns. for one of the columns, i need to take the average for only
> those fields where the content !='6' (with 6 being a Flag telling me
> to exclude it). so i came up with the following
> select statement, which returns no mysql_error, but does not return the
> correct figure for the AVG:
>
> SELECT AVG(q1), AVG(q2), AVG(q3), AVG(q4), AVG(q5), AVG(q6), AVG(q7),
> AVG(q8), AVG(q11), AVG(q16), AVG(q18!='6') FROM hr_form2 WHERE Answer !=
> 0
>
> from this i should get a return of something like this, since NONE of my
> fields in column q18 contain a '6':
>
> AVG(q18) = 2.7143
>
> instead, i get this:
>
> AVG(q18) = 1.0000
>
> could someone please tell me--is there a logical error in my thinking
> here? i want to select the AVG from q18 ONLY for those fields where
> q18!=6 : for those fields where q18=6, I just want it left out of the
> AVG.
>
> there's definitely a better way to formulate my sql, right?
>
> muchos gracias,
>
> tom 

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