[RESOLVED] Harumph. Formula's not working

[duckduckgoose]

I'm trying to write a script that says, if $n can be divided by 3 with no remainder, then do X, otherwise, do Z. But, no matter what, it does Z. Note that the $row['count(*)'] equals 24. I've run the select statement in the SQL window, and it's most definitely 24.

Here's the script:

        $queryx = "SELECT count(*) FROM mytable WHERE e='".$_GET['e']."'";
        $resultx = mysql_query($queryx);

    while($row = mysql_fetch_assoc($resultx)) {
            $ec2 = $row['count(*)'] / 3;

if(($ec2%3)==0) {

echo "ok";

} else {

$ec3 = floor($ec2);

echo "not ok";

}
}

[Crowly]

This code doesnt make much sence to me

$queryx = "SELECT count(*) FROM mytable WHERE e='".$_GET['e']."'"; 
        $resultx = mysql_query($queryx); 

    while($row = mysql_fetch_assoc($resultx)) { 
            $ec2 = $row['count(*)'] / 3;

Since the select will only return one row there is no need for the while. You can do it like this instead:

$queryx = "SELECT count(*) n FROM mytable WHERE e='".$_GET['e']."'"; 
        $resultx = mysql_query($queryx); 
        $row = mysql_fetch_assoc($resultx) ;
        $ec2 = $row['n'] / 3;

or

$queryx = "SELECT somefield FROM mytable WHERE e='".$_GET['e']."'"; 
        $resultx = mysql_query($queryx); 
        $n = mysql_num_rows($resultx);
        $ec2 = $n / 3;

[Piranha]

Use another name for the count parameter, like below:

 $queryx = "SELECT count(*) as nr FROM mytable WHERE e='".$_GET['e']."'"; 

Then further down use nr instead of count(*):

$ec2 = $row['nr'] / 3;

Another good idea is to echo the variable you have problem with in the page, then you will get to know if it really is 24 here or not.

[duckduckgoose]

Thanks. I figured out the problem shortly after I posted. My formulas were all out of order. :/