使用SQL从数据库中选择数据

Question

我有一个sql子选择来列出有多少个订单及其总价值：

SELECT u.name,
  (SELECT COUNT(*) FROM orders o WHERE o.user_id=u.id
  ) AS order_count,
  (SELECT SUM(oi.quantity * p.price)
  FROM orders            AS o
  INNER JOIN order_items AS oi
  ON oi.order_id = o.id
  INNER JOIN products AS p
  ON p.id        = oi.product_id
  WHERE o.user_id=u.id
  ) AS total_price
FROM users u
ORDER BY total_price DESC

任何想法，我如何做到这一点而无需子选择？

Answer 1

您可以使用Left Join

SELECT u.NAME,
       AS order_count,
       total_price
FROM   users u
       LEFT JOIN (SELECT Count(*) AS order_count,
                         user_id
                  FROM   orders
                  GROUP  BY user_id) p
              ON o.user_id = u.id
       LEFT JOIN (SELECT Sum(oi.quantity * p.price) AS total_price,
                         o.user_id
                  FROM   orders AS o
                         INNER JOIN order_items AS oi
                                 ON oi.order_id = o.id
                         INNER JOIN products AS p
                                 ON p.id = oi.product_id
                  GROUP  BY o.user_id) so
              ON so.user_id = u.id
ORDER  BY total_price DESC 

Answer 2

您可以将相关的子查询（O（N ^ 2）个性能）转换为LEFT JOIN

SELECT u.name,
  o.order_count,
  o2.total_price
FROM users u left join (
  SELECT user_id, COUNT(*) order_count
  FROM orders
  GROUP BY user_id
) o
on u.id = o.user_id
left join (
  SELECT o.user_id, SUM(oi.quantity * p.price) total_price
  FROM orders o
  INNER JOIN order_items oi
  ON oi.order_id = o.id
  INNER JOIN products p
  ON p.id        = oi.product_id
  GROUP BY o.user_id
) o2 on u.id = o2.user_id
ORDER BY total_price DESC;

Answer 3

这是带有一个子查询的相同查询（注意，未经语法测试，但您知道了）：

select u.name, count(*) orderCount, sum(totalOrder) total order
  from users u
         join (select userid, o.id, sum(oi.quantity * p.price) totalorder
                from orders o
                    join order_items as oi on oi.order_id = o.orderid
                    join products p on oi.product_id = p.id
                group by userid, o.id) x
on u.id = x.user_id
group by u.name