我有两张桌子
顺序:
order_id | order_date | order_by
----------------------------------
1 2018-01-05 suraj
2 2018-01-06 sumanta
order_detail:
order_detail_id | order_id | item_id | item_cost
-----------------------------------------------------
1 1 23 100
2 1 27 200
3 2 16 300
4 2 17 100
5 2 10 400
如何从mysql
中的上表中获取此数据order_id | order_date | order_by | no_of_item | order_cost
----------------------------------------------------------------
1 2018-01-05 suraj 2 300
2 2018-01-06 sumanta 3 700
答案 0 :(得分:1)
加入两个表,并与GROUP BY汇总。
SELECT t1.order_id, t1.order_date, t1.order_by, count(*) as no_of_items, sum(t2.item_cost) as order_cost
FROM order t1
INNER JOIN order_detail t2
ON t1.order_id = t2.order_id
GROUP BY t1.order_id, t1.order_date, t1.order_by;
答案 1 :(得分:0)
SELECT o.order, o.order_date, o.order_by, COUNT(*), SUM(od.item_cost)
FROM order o JOIN order_detail od ON(o.order_id=od.order_id)
GROUP BY o.order_id, o.order_date, o.order_by;
我觉得我在做某人的家庭作业
答案 2 :(得分:0)
使用SQL查询。
SELECT o.order_id, o.order_date, o.order_by,
COUNT(*) as no_of_item, SUM(od.item_cost) as order_cost
FROM order o INNER JOIN order_detial od ON o.order_id = od.order_id
GROUP BY o.order_id, o.order_date, o.order_by;