Question

So I wrote this earlier (in php), but everytime I try echo $test”, I just get back resource id 5. Does anyone know how to actually print out the mysql query from the variable?

$dave= mysql_query("SELECT order_date, no_of_items, shipping_charge, SUM(total_order_amount) as test FROM `orders` WHERE DATE(`order_date`) = DATE(NOW()) GROUP BY DATE(`order_date`)") or die(mysql_error());
print $dave;

Answer 1

This will print out the query:

$query = "SELECT order_date, no_of_items, shipping_charge, SUM(total_order_amount) as test FROM `orders` WHERE DATE(`order_date`) = DATE(NOW()) GROUP BY DATE(`order_date`)";

$dave= mysql_query($query) or die(mysql_error());
print $query;

This will print out the results:

$query = "SELECT order_date, no_of_items, shipping_charge, SUM(total_order_amount) as test FROM `orders` WHERE DATE(`order_date`) = DATE(NOW()) GROUP BY DATE(`order_date`)";

$dave= mysql_query($query) or die(mysql_error());

while($row = mysql_fetch_assoc($dave)){
    foreach($row as $cname => $cvalue){
        print "$cname: $cvalue\t";
    }
    print "\r\n";
}