以下查询no.1和查询no.2正在处理每一行.. 1.
string = "a=1&b=2&c=3&d=4&nbr=1234567890"
2
SELECT app_user.au_userid, SUM(order_detail.od_quantity), (order_detail.od_quantity*SUM(order_detail.od_subtotal)) AS total_amount FROM app_user JOIN orders ON app_user.au_id = orders.o_au_id JOIN order_detail ON orders.o_id = order_detail.od_order_id GROUP BY app_user.au_userid
以下查询no.3不起作用.. 3。
SELECT COUNT (app_user.au_id) FROM app_user JOIN orders ON app_user.au_id = orders.o_au_id GROUP BY app_user.au_id
答案 0 :(得分:0)
最简单的方法是使用COUNT(distinct)
:
SELECT o.o_au_id, COUNT (DISTINCT o.o_id) AS num_orders,
SUM(od.od_quantity),
SUM(od.od_quantity * od.od_subtotal) AS total_amount
FROM orders o JOIN
order_detail od
ON o.o_id = od.od_order_id
GROUP BY o.o_au_id;
注意:
app_users
,因为您在订单表中拥有用户ID。count(distinct)
。