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Printing result of mysql query from variable

Tags:

php

mysql

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;
like image 482
Louis Stephens Avatar asked Sep 22 '11 18:09

Louis Stephens


2 Answers

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";
}
like image 169
Jeremy Holovacs Avatar answered Oct 16 '22 09:10

Jeremy Holovacs


well you are returning an array of items from the database. so you need something like this.

   $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());


while ($row = mysql_fetch_assoc($dave)) {
echo $row['order_date'];
echo $row['no_of_items'];
echo $row['shipping_charge'];
echo $row['test '];
}
like image 21
Lucas Avatar answered Oct 16 '22 08:10

Lucas