[BigTime]

Im trying to gather a sum of records that match a certain criteria, then return the number of records that match only. This would basically be for a win loss column for a jr football league.

Ive messed around all day and still cant get it to work.

My latest rendition is the following code, but it is returning an error on the line that begins with $num_rows -- Exactly it says:
mysql_num_rows(): supplied argument is not a valid MySQL result


I just cant figure out what is going on and Im pulling my hair out....for sure there are 2 records that match the select criteria....

Can anyone that is a bit more experienced in this help?

<.?php

$division = "Featherweight";
$league = "Sunday";
$hometeam = "Fox Lake";

$result = mysql_query("SELECT * FROM `schedules` WHERE `league` = '$league' AND `division` = '$division' AND `hometeam` = '$hometeam' AND `homescore` > `awayscore` FROM schedules");
$num_rows = mysql_num_rows($result);

echo "$num_rows Rows\n";

?.>

[NeverMind]

you shouldn't have 2 FROMs in the query!
use this query instead:

Code:

SELECT * FROM schedules WHERE league = '$league' AND division = '$division' AND hometeam = '$hometeam' AND homescore > awayscore

[BigTime]

Hehe

Well that just how convoluted my thought process got yesterday I guess.....FROM started out at the beginning and then ended up at the end, and neither got deleted.

HOWeva......

Still same error

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource Calling from the line that is the beginning of $num_Rows


<?php

$division = "Featherweight";
$league = "Sunday";
$hometeam = "Fox Lake";

$result = mysql_query("SELECT * FROM schedules WHERE league = '$league' AND division = '$division' AND hometeam = '$hometeam' AND homescore > awayscore");
$num_rows = mysql_num_rows($result);

echo "$num_rows Wins\n";

?>

[NeverMind]

then try changing this:

PHP Code:

$result = mysql_query("SELECT * FROM schedules WHERE league = '$league' AND division = '$division' AND hometeam = '$hometeam' AND homescore > awayscore"); 

to this:

PHP Code:

$result = mysql_query("SELECT * FROM schedules WHERE league = '$league' AND division = '$division' AND hometeam = '$hometeam' AND homescore > awayscore")or

die(mysql_error()); 

and see what is the error message that appears and tell us ..

[BigTime]

hmmm....

Well now Im all embarassed -- It says no Database selected......I'm connecting the same way I normally do...but I'm sure there was something simple here Im overlooking.......

Full on code:

<?
$usr = "username";
$pwd = "password";
$db = "database";
$host = "localhost";

# connect to database
$cid = mysql_connect($host,$usr,$pwd);
if (!$cid) { echo("ERROR: " . mysql_error() . "\n"); }

$division = "Featherweight";
$league = "Sunday";
$hometeam = "Fox Lake";

$result = mysql_query("SELECT * FROM schedules WHERE league = '$league' AND division = '$division' AND hometeam = '$hometeam' AND homescore > awayscore")or
die(mysql_error());

$num_rows = mysql_num_rows($result);

echo "$num_rows Wins\n";

?>


Working Now and I changed it to:

<?
$dbuser="userinfo";
$dbpass="password";
$dbname="database name"; //the name of the database
$cid = mysql_connect("localhost", $dbuser, $dbpass)
or die("Connection Failure to Database");
mysql_select_db($dbname, $cid) or die ($dbname . " Database not found." . $dbuser);


$division = "Featherweight";
$league = "Sunday";
$hometeam = "Fox Lake";

$result = mysql_query("SELECT * FROM schedules WHERE league = '$league' AND division = '$division' AND hometeam = '$hometeam' AND homescore > awayscore", $cid)or
die(mysql_error());

$num_rows = mysql_num_rows($result);

echo "$num_rows Wins\n";
mysql_close($cid);
?>


Thank you Nevermind very much for getting the juices flowing and helping me sort this out!!!