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
You can name the pseudo-column by doing select AVG(age) as avgage from ...
and then accessing it as avgage.
- Tim
http://www.phptemplates.org
- Original Message -
From: Jeff Lewis [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Tuesday, July 10, 2001 11:27 AM
Subject: [PHP] Average
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
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
2001 16:54
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
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
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