按日期显示求和结果组mysqli

Question

我需要显示总价为ORDER BY order_date的表时出现错误的结果，这是一个问题

我有一个数据库：

- order_id
- order_date
- order_price
- order_quantity

我试过了 ：

$sql = mysqli_query($con,"SELECT *, 
                                 SUM(order_q) * SUM(order_price) as total 
                          FROM o_orders_f 
                          GROUP BY order_id");

但是这段代码将汇总所有结果，并给我错误的答案

$sql = mysqli_query($con, "SELECT *, 
                                  SUM(order_quantity) * SUM(order_price) as total 
                           FROM o_orders_f 
                           GROUP BY order_id");
while ($r = mysqli_fetch_array($sql)) {
   $date = $r['order_date'];
   $total = $r['total'];
   echo $total;     
}

我正在寻找的结果是这样的：

7 Jul, 2019 - total: ($r['quantity']* $r['price']) = 72933 
8 Jul, 2019 - total: ($r['quantity']* $r['price']) = 453
9 Jul, 2019 - total: ($r['quantity']* $r['price']) = 72

Answer 1

我在您的查询中看到2个错误：

1. 当您应该按order_date分组时，您就是按order_id分组。
2. 当在查询乘以值，你应该SUM操作的结果，但你正在总结它的因素。

也就是说，您的查询应为：

SELECT
    *,
    SUM(order_quantity * order_price) as total 
FROM 
    o_orders_f 
GROUP BY 
    order_date