This doesn't work:
$age_result = mysql_query("select AVG(age) as avgage FROM bat_rost WHERE
ownerID = '$teamID'");
$row = mysql_fetch_object($age_result);
$age=$avgage;
echo "Average age - ".$teamID.$avgage;
Neither does this:
$age_result = mysql_query("select AVG(age) as avgage FROM bat_rost WHERE
ownerID = '$teamID'");
$res = mysql_fetch_row($age_result)
$age=$res[0];
echo "Average age - ".$teamID.$avgage;
> -----Original Message-----
> From: Remo Pini [mailto:[EMAIL PROTECTED]]
> Sent: Tuesday, July 10, 2001 11:41 AM
> To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
> Subject: RE: [PHP] Average of column...
>
>
> if you do a
>
> $res = mysql_fetch_row($age_result)
>
> $res[0] will be the value of AVG(age).
>
>
> > -----Original Message-----
> > From: Jeff Lewis [mailto:[EMAIL PROTECTED]]
> > Sent: Tuesday, July 10, 2001 5:27 PM
> > To: [EMAIL PROTECTED]
> > Subject: [PHP] Average of column...
> >
> >
> > I am trying to obtain the average age for a few teams in my database.
> >
> > Am using the below code:
> >
> > $age_result = mysql_query("select AVG(age) FROM bat_rost WHERE ownerID =
> > '$teamID'");
> > while(($row = mysql_fetch_object($age_result))){
> > $age=$row->age;
> > echo "Average age - ".$teamID.$age;}
> >
> > How can I get the average age from this?
> >
> > Jeff
> >
> >
> > --
> > PHP General Mailing List (http://www.php.net/)
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> >
> >
>
>
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