[Resolved] problem with an array query

[ejwf]

Hello
I have data in the mysql database that I know is there but cannot get the code right to get it out........

code is

$today = date('Y-m-d');

$query = "SELECT Date, Ref FROM tblDate WHERE First = $today";
$result = mysql_query($query)
	or die(mysql_error());
while ($row = mysql_fetch_array($result));
$Ref[] = $row["Ref"];
$Date[] = $row["Date"];

print_r($Date);
print_r($Ref);

when I test this the data that should come out (because I put the data in the database specifically to test this) is
Array ( [0] =>8, [1]=>7 ) Array ( [0] =>1, [1] =>1 )

but instead I am getting
Array ( [0] => ) Array ( [0] => )

any ideas what I am getting wrong?

TIA

[laserlight]

What if you change the while loop into just:
$row = mysql_fetch_array($result);

[devinemke]

your syntax is a bit off. also use [man]mysql_num_rows[/man] to see if ther are any records matching your criteria:

$today = date('Y-m-d'); 

$query = "SELECT Date, Ref FROM tblDate WHERE First = '$today'";
$result = mysql_query($query) or die(mysql_error());
if (mysql_num_rows($result))
{
	while ($row = mysql_fetch_array($result))
	{ 
		$Date[] = $row['Date'];
		$Ref[] = $row['Ref'];
	} 

print_r($Date);
print_r($Ref);
}
else
{
	echo 'no records found matching that criteria';
}

[ejwf]

laserlight -
if I do

$today = date('Y-m-d');
$query = "SELECT Date, Ref FROM tblDate WHERE First = $today";
$result = mysql_query($query)
	or die(mysql_error());
$row = mysql_fetch_array($result);
$Ref = $row["Ref"];
$Date = $row["Date"];

echo $Ref;
echo $Date;

I get nothing - no output at all

devinemke -
following your code I get

Array ( [0] => 7 [1] => 8 ) Array ( [0] => 1 [1] => 2 )

which are the results I was expecting
thank you.
I think I can take it forward from here

thank you both for your time