UNION & ORDER 两个表里面 Common Table Expression

Question

我在 SQL 存储过程中有一个 CTE，它 UNION 从两个数据库获取值 - 这些值是客户编号和该客户的最后订单日期。

这里是原文SQL-

;WITH CTE_last_order_date AS
(
SELECT c1.customer ,MAX(s2.dt_created) AS last_order_date
FROM customers c1 WITH (NOLOCK)

LEFT JOIN archive_orders s2 WITH (NOLOCK)
ON c1.customer = s2.customer

GROUP BY c1.customer

UNION ALL

SELECT c1.customer ,MAX(s1.dt_created) AS last_order_date
FROM customers c1 WITH (NOLOCK)

LEFT JOIN orders s1 WITH (NOLOCK)
ON c1.customer = s1.customer

GROUP BY c1.customer
)

示例结果：

customer,    last_order_date
CF122595,    2011-11-15 15:30:22.000
CF122595,    2016-08-15 10:01:51.230

(2 row(s) affected)

这显然不适用 UNION 不同的记录规则，因为日期值不匹配，这意味着 SQL 从 both 返回最大值表（即最终记录集不明确）

为了尝试解决这个问题，我尝试了从 this question 借用的另一种方法并实现了分组：

;WITH CTE_last_order_date AS
(
SELECT max(last_order_date) as 'last_order_date', customer
FROM (
SELECT distinct cust.customer, max(s2.dt_created) AS last_order_date, '2' AS 'group'
FROM customers c1 WITH (NOLOCK)

LEFT JOIN archive_orders s2 WITH (NOLOCK)
ON c1.customer = s2.customer

GROUP BY c1.customer

UNION 

SELECT distinct c1.customer, max(sord.dt_created) AS last_order_date, '1' AS 'group'
FROM customers c1 WITH (NOLOCK)

LEFT JOIN orders s1 WITH (NOLOCK)
ON cust.customer = sord.customer

GROUP BY
   c1.customer
   ) AS t
GROUP  BY customer
ORDER  BY MIN('group'), customer
)

示例结果：

customer,    last_order_date
CF122595,    2016-08-15 10:01:51.230

(1 row(s) affected)

这具有工作正常的区别 (hah)，直到在 Common Table 表达式中进入阻止 ORDER BY 的规则，这是选择最低组所必需的（这将意味着实时订单（第 1 组），其日期需要优先于存档（第 2 组））。

The ORDER BY clause is invalid in views, inline functions, derived tables, subqueries, and common table expressions, unless TOP or FOR XML is also specified.

感谢所有帮助或想法。

Answer 1

与其分组，然后合并，然后再次分组，为什么不合并订单表并从那里开始工作：

SELECT c1.customer ,MAX(s2.dt_created) AS last_order_date
FROM customers c1
INNER JOIN (select customer, dt_created from archive_orders
union all select customer, dt_created from orders) s2
ON c1.customer = s2.customer
GROUP BY c1.customer

记住，在 SQL 中，你的工作是告诉系统 你想要什么 ，而不是 要遵循什么 steps/procedure这些结果。从逻辑上讲，上面描述了我们想要的 - 我们想要每个客户订单的最后订单日期，我们不关心这是存档订单还是非存档订单。

由于我们要在 GROUP BY 行为期间将订单信息减少到一行（每个客户），所以我们也不需要UNION 删除重复项所以我切换到 UNION ALL.

（我承认，此时我真的看不出 ORDER BY 应该添加什么，所以我没有尝试将其包含在这里。如果这将进入CTE，那就反思一下，CTE和表、视图一样，没有inherent顺序，唯一影响结果行顺序的ORDER BY子句是应用的到 outermost/final SELECT)

orders 优先于 archived_orders：

;With CTE1 as (
    SELECT c1.customer,group,MAX(s2.dt_created) as MaxInGroup
    FROM customers c1
    INNER JOIN (select customer, dt_created,2 as group from archive_orders
    union all select customer, dt_created,1 from orders) s2
    ON c1.customer = s2.customer
    GROUP BY c1.customer,group
), CTE2 as (
    SELECT *,ROW_NUMBER() OVER (PARTITION BY customer ORDER BY group) as rn
    from CTE2
)
select * from CTE2 where rn = 1

Answer 2

另一种方法可能是仅从我们没有当前档案的档案 table 中获取客户。类似于：

WITH CurrentLastOrders(customer, last_order_date) AS    -- Get current last orders
(
    SELECT o.customer, max(o.dt_created) AS last_order_date
    FROM orders s WITH (NOLOCK) ON c.customer = o.customer
    GROUP BY o.customer
),
ArchiveLastOrders(customer, last_order_date) AS -- Get archived last orders where customer does not have a current order
(
    SELECT o.customer, max(o.dt_created) AS last_order_date
    FROM archive_orders o WITH (NOLOCK)
    WHERE NOT EXISTS ( SELECT *
                        FROM CurrentLastOrders lo
                        WHERE o.customer = lo.customer)
    GROUP BY o.customer
),
AllLastOrders(customer, last_order_date) AS -- All customers with orders
(
    SELECT customer, last_order_date
    FROM CurrentLastOrders
    UNION ALL
    SELECT customer, last_order_date
    FROM ArchiveLastOrders
)
AllLastOrdersPlusCustomersWithNoOrders(customer, last_order_date) AS    -- All customerswith latest order if they have one
(
    SELECT customer, last_order_date
    FROM AllLastOrders
    UNION ALL
    SELECT customer, null
    FROM customers c WITH (NOLOCK)
    WHERE NOT EXISTS ( SELECT *
                        FROM AllLastOrders lo
                        WHERE c.customer = lo.customer)
)

Answer 3

我不会尝试嵌套 SQL 来获得不同的结果集，这与在两个联合查询中按客户分组的逻辑相同。如果你想要一个不同的有序集，你可以在 CTE 之外做到这一点

怎么样：

;WITH CTE_last_order_date AS
(
   SELECT c1.customer ,s2.dt_created AS last_order_date, '2' AS 'group'
   FROM customers c1 WITH (NOLOCK)
   LEFT JOIN archive_orders s2 WITH (NOLOCK) ON c1.customer = s2.customer

   UNION ALL

   SELECT c1.customer ,s1.dt_created AS last_order_date, '1' AS 'group'
   FROM customers c1 WITH (NOLOCK)
   LEFT JOIN orders s1 WITH (NOLOCK) ON c1.customer = s1.customer

)
SELECT customer, MAX(last_order_date)
FROM CTE_last_order_date
GROUP BY customer 
ORDER BY MIN('group'), customer

Answer 4

如果将所有可能的行合并在一起，然后计算 row_number，按客户分区并按 'group' 排序，然后 last_order_date 降序，然后可以 select 所有row=1 给每个客户'top 1'

;WITH CTE_last_order_date AS
(
SELECT max(last_order_date) as 'last_order_date', customer
FROM (
SELECT distinct cust.customer, max(s2.dt_created) AS last_order_date, '2' AS 'group'
FROM customers c1 WITH (NOLOCK)

LEFT JOIN archive_orders s2 WITH (NOLOCK)
ON c1.customer = s2.customer

GROUP BY c1.customer

UNION 

SELECT distinct c1.customer, max(sord.dt_created) AS last_order_date, '1' AS 'group'
FROM customers c1 WITH (NOLOCK)

LEFT JOIN orders s1 WITH (NOLOCK)
ON cust.customer = sord.customer

GROUP BY
   c1.customer
   ) AS t
GROUP  BY customer

)
,   --row_number below is 'per customer' and can be used to make rn=1 the top 1 for each customerid
ROWN AS (SELECT Customer,last_order_date,[group], row_number() OVER(partition by customer order by [group] ASC, sord.dt_created DESC) AS RN)
SELECT * FROM Rown WHERE Rown.rn = 1

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