检索没有特定项目的订单号

Question

我有一张包含一些冰淇淋订单数据的表格如下：

id  order_number  item_name
1   1             cake cone
2   1             sprinkles
3   2             sugar cone
4   3             semi-sweet morsels
5   3             sprinkles

（这比真实表更简单，但存在相关列。）

基本上我想要做的是生成一个不包含特定项目（糖锥和蛋糕锥）的订单号列表。

我在这一点上尝试的似乎有效，但我想知道是否有更好的方法：

select a.order_number from 
(select order_number from orders where (item_name != 'sugar cone' and item_name != 'cake cone') group by order_number)a
left join
(select order_number from orders where (item_name = 'sugar cone' or item_name = 'cake cone') group by order_number)b
on a.order_number = b.order_number
where b.order_number is null;

这将返回我期望的order_numbers（在这种情况下，只有3阶），但对我来说似乎很笨拙。

你有更好的方法吗？

Answer 1

使用LEFT JOIN / IS NULL：

   SELECT x.order_number
     FROM ORDERS x
LEFT JOIN ORDERS y ON y.order_number = x.order_number
                  AND y.item_name IN ('sprinkles', 'sugar cone')
    WHERE y.id IS NULL

使用NOT EXISTS

   SELECT x.order_number
     FROM ORDERS x
    WHERE NOT EXISTS(SELECT NULL
                       FROM ORDERS y
                      WHERE y.item_name IN ('sprinkles', 'sugar cone')
                        AND y.order_number = x.order_number)

使用NOT IN

   SELECT x.order_number
     FROM ORDERS x
    WHERE x.order_number NOT IN(SELECT y.order_number
                        FROM ORDERS y
                      WHERE y.item_name IN ('sprinkles', 'sugar cone'))

摘要

在MySQL上，if the column(s) being joined on are not nullable - LEFT JOIN/IS NULL is the fastest/most efficient。否则，NOT EXISTS & NOT IN are better choices。

结论

由于加入了不能为NULL的列，因此LEFT JOIN / IS为最佳选择。

Answer 2

使用EXISTS子句将比使用IN子句提供更好的性能。

SELECT a.order_number
  FROM orders a
 WHERE NOT EXISTS (SELECT 1
                     FROM orders b
                    WHERE b.item_name IN ('sugar cone', 'cake cone')
                          AND b.order_number = a.order_number);

Answer 3

select order_number
from orders 
where order_number not in (
    select order_number
    from orders 
    where item_name in ('sprinkles', 'sugar cone')
)

Answer 4

select distinct order_number from orders where order_number not in 
    (select order_number from orders where item_name in ('sugar cone', 'cake cone'))