从变量打印mysql查询的结果

Question

所以我之前写的(在php中),但每次我尝试echo $ test",我只是回到资源id 5.有谁知道如何从变量中实际打印出mysql查询？

$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

这将打印出查询:

$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;

这将打印出结果:

$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";
}

Answer 2

那么你从数据库返回一个项目数组.所以你需要这样的东西.

   $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 '];
}