聚合函数和子查询的语法问题

2024-05-05 • 问答

我正在尝试查询3表数据库Orders，客户和销售员。为了找到有关销售人员最近工作表现的更多信息。所需的类别是id，age，amout> 1000的订单号，数量> 500（y / n）的订单，服务的唯一客户数，自上次订单以来的天数以及最后一次下单的金额。

我的代码如下，也可以在链接中查看：

SELECT  o.salesperson_id
    s.Age,(CASE WHEN o.Amount > 500 THEN 'Yes' ELSE 'No' END) AS 'Had Order Bigger Than 500?',(datediff(day,o.order_date,GETDATE())) AS 'Days Since Last Order',o.Amount as 'Last Order Amount'
    FROM Orders o INNER JOIN Salesperson s ON o.salesperson_id = s.ID
      WHERE o.Amount IN
      (SELECT o.Amount WHERE o.order_date = MAX (o.order_date) FROM Orders)

    GROUP BY o.salesperson_id 
    HAVING (COUNT (DISTINCT o.cust_id)) AS 'Number of Unique Customers Served',(SUM(CASE WHEN o.Amount > 1000 THEN 1 ELSE 0 END)) AS 'Number of Orders Bigger Than 1000'

    ORDER BY o.salesperson_id;

http://sqlfiddle.com/#!18/4fc5c/69

自动调试器说问题出在我的FROM上，但是我很傻

关键字“ FROM”附近的语法不正确

我的预期结果将是这样：

+----------------+-----------------+-----------------------------------+-----------------------------------+-----------------------------------+-----------------------+-------------------+
| Salesperson ID | Salesperson Age | Number of Orders Bigger Than 1000 | Had Order Bigger Than 500? (y/n)  | Number of Unique Customers Served | Days Since Last Order | Last Order Amount |
+----------------+-----------------+-----------------------------------+-----------------------------------+-----------------------------------+-----------------------+-------------------+
|              1 |              61 |                                 0 | N                                 |                                 1 |                       |               460 |
|              2 |              34 |                                 1 | Y                                 |                                 2 |                       |              2400 |
|              8 |              57 |                                 1 | y                                 |                                 1 |                       |              1800 |
+----------------+-----------------+-----------------------------------+-----------------------------------+-----------------------------------+-----------------------+-------------------+

将非常感谢收到的所有援助。

您的查询syntax完全错误。请尝试这个。

FIDDLE DEMO

SELECT  o.salesperson_id,s.Age,(SUM(CASE WHEN o.Amount > 1000 THEN 1 ELSE 0 END)) AS 'Number of Orders Bigger Than 1000',(CASE WHEN o.Amount > 500 THEN 'Yes' ELSE 'No' END) AS 'Had Order Bigger Than 500?',(COUNT (o.cust_id)) AS 'Number of Unique Customers Served',(datediff(day,o.order_date,GETDATE())) AS 'Days Since Last Order',o.Amount as 'Last Order Amount'  
FROM Orders o INNER JOIN Salesperson s ON o.salesperson_id = s.ID
WHERE o.Amount IN
( 
  SELECT Amount 
  FROM Orders OO
  WHERE order_date IN (SELECT MAX(order_date) FROM Orders GROUP BY salesperson_id)
)
GROUP BY o.salesperson_id,o.Amount,GETDATE()))
ORDER BY o.salesperson_id;

加入销售人员和订单，按销售人员分组并使用条件汇总：

select
  s.ID [Salesperson ID],s.Age [Salesperson Age],sum(case when o.Amount > 1000 then 1 else 0 end) [Number of Orders Bigger Than 1000],case when sum(case when o.Amount > 500 then 1 else 0 end) > 0 then 'Yes' else 'No' end [Had Order Bigger Than 500? (y/n)],count(distinct cust_id) [Number of Unique Customers Served],datediff(day,max(o.order_date),GETDATE()) [Days Since Last Order],(select Amount from Orders where salesperson_id = s.ID and order_date = max(o.order_date))[Last Order Amount]
from Salesperson s inner join Orders o
on o.salesperson_id = s.ID
group by s.ID,s.Age
order by s.ID

请参见demo。
结果：

> Salesperson ID | Salesperson Age | Number of Orders Bigger Than 1000 | Had Order Bigger Than 500? (y/n) | Number of Unique Customers Served | Days Since Last Order | Last Order Amount
> -------------: | --------------: | --------------------------------: | :------------------------------- | --------------------------------: | --------------------: | ----------------:
>              1 |              61 |                                 0 | No                               |                                 1 |                  8883 |               460
>              2 |              34 |                                 1 | Yes                              |                                 2 |                  7953 |              2400
>              8 |              57 |                                 1 | Yes                              |                                 1 |                  7587 |              1800

聚合函数和子查询的语法问题

leiximo 回答：聚合函数和子查询的语法问题

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