hm, not too surprising, no colon delimiter...
$age_result = mysql_query("select AVG(age) FROM bat_rost WHERE ownerID =
'$teamID'");
$res = mysql_fetch_row($age_result);
$age = $res[0];
echo "Average age - $teamID$age";
should work actually... btw, if teamID is numeric, you have to omit the '
thingys...
> -----Original Message-----
> From: Jeff Lewis [mailto:[EMAIL PROTECTED]]
> Sent: Tuesday, July 10, 2001 5:54 PM
> To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
> Subject: RE: [PHP] Average of column...
>
>
> 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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