Help with Query

[james0816]

ok...this code works in mysql....but not displaying properly in PHP.

$result3 = mysql_query("select avg(s) as avs, avg(f) as avf from results where dcode='1' and yr='2005'", $connection);

echo mysql_result($result3,"avs");

This statement will display avs correctly. However, If I change it fo avf....it still displays the result from avs????

i am soooo confused. Can someone shed some light on this?

thx

 

[kenrbnsn]

It would be better (IMHO) to write:

$q = "select avg(s) as avs, avg(f) as avf from results where dcode='1' and yr='2005'";
$rs = @mysql_query($q) or die('Problem with query. ' . mysql_error());
$rw = @mysql_fetch_assoc($rs);
echo 'avs: ' . $rw['avs'] . "<br>\n";
echo 'avf: ' . $rw['avf'] . "<br>\n";

It's much clearer as to what the code is doing.

Ken

 

[james0816]

i have it working now but even more confused.

i wound up coding it like this:

echo mysql_result($result3,0,0) for avs
echo mysql_result($result3,0,1) for avf

it displayed fine that way....but why not by name??

 

[LTeeple]

You are missing a parameter for the PHP function mysql_result.

mysql_result(resource result,int row [,mixed field])

See PHP Manual - mysql_result()


Cheers!
Laura