SQL Server 中的水平 DISTINCT COUNT 或不计算重复项的 COUNT IF

Question

我已经过度简化了以下示例，但请注意，我确实需要我所描述的结果。

我有一个表格，其中包含有关客户的信息，包括他们是如何被推荐的。 例如MyTable（编辑：我的过度简化导致答案不起作用。它们是很好的答案，但不能解决我的问题。为了清楚起见，我添加了最后一块拼图）：

STORE    CUS_ID REFERRED    MONEYMADE  
 1        11        RADIO         10.00
 1        15        WALKIN        20.00
 1        11        RADIO         30.00
 2        12        RADIO         40.00
 2        12        RADIO         50.00
 3        13        WALKIN        60.00
 3        14        WALKIN        70.00

我想统计按特定类型推荐的 DISTINCT 客户的数量，并按商店对他们进行分组。我可以做到以下几点：

SELECT 
   STORE, COUNT(DISTINCT CUS_ID) AS COUNTEM, 
   REFERRED, SUM(MONEYMADE) 
FROM MyTable 
GROUP BY REFERRED, STORE

这给出了以下结果：

STORE    COUNTEM    REFERRED    MONEYMADE
 1        1         RADIO        40.00
 1        1         WALKIN       20.00
 2        1         RADIO        90.00
 3        2         WALKIN       130.00

但是，我真的需要这样的结果：

STORE    RADIO  RADIO_MONEY   WALKIN   WALKIN_MONEY
 1        1      40.00         1         20.00
 2        1      90.00         0         0.00
 3        0      0.00          2         130.00

我尝试使用 SUM 和 CASE 进行查询，但它计算了重复项。例如：

SELECT 
   STORE, 
   SUM (CASE REFERRED WHEN 'RADIO' THEN 1 ELSE 0 END) AS RADIO,  
   SUM (CASE REFERRED WHEN 'RADIO' THEN MONEYMADE ELSE 0 END) AS RADIO_MONEY,  
   SUM (CASE REFERRED WHEN 'WALKIN' THEN 1 ELSE 0 END) AS WALKIN
   SUM (CASE REFERRED WHEN 'WALKIN' THEN MONEYMADE ELSE 0 END) AS WALKIN_MONEY, 
FROM MyTable
GROUP BY STORE

返回这个不正确的结果：

STORE    RADIO  RADIO_MONEY   WALKIN   WALKIN_MONEY
 1        2      40.00         1         20.00
 2        2      90.00         0         0.00
 3        0      0.00          2         130.00

有没有办法解决这个问题？我已经搜索了“count if”类型的函数，但我唯一找到的是 sum - case 方法。

Answer 1

EDIT 因为您需要对另一个字段求​​和。你应该在 SUM/CASE 之前做总和

自带样本数据

--TEST DATA
DECLARE @MyTable table 
(STORE int,  CUST_ID int , Referred varchar(6) ,MONEYMADE   Money )

INSERT INTO @MyTable VALUES  
(1,        11,        'RADIO' ,        10.00)
INSERT INTO @MyTable VALUES  
(1,        15,        'WALKIN',        20.00)
INSERT INTO @MyTable VALUES 
(1,        11,        'RADIO' ,       30.00)
INSERT INTO @MyTable VALUES 
(2,        12,        'RADIO' ,        40.00)
INSERT INTO @MyTable VALUES 
(2,        12,        'RADIO' ,       50.00)
INSERT INTO @MyTable VALUES 
(3,        13,        'WALKIN',        60.00)
INSERT INTO @MyTable VALUES 
(3,        14,        'WALKIN',        70.00)

--使用枢轴

  SELECT 
    pk.STORE,
    pk.RADIO,
    pk.WALKIN,
    ISNULL(pv.RADIO,0) as RADIO_MONEY,
    ISNULL(pv.WALKIN,0)as WALKIN_MONEY

 FROM
 (SELECT  STORE,   
          REFERRED ,
          MONEYMADE   
  FROM  @MyTable) p
          PIVOT (SUM(MONEYMADE) FOR REFERRED in (WALKIN, RADIO)) as  pv
  INNER JOIN         
 ( SELECT  DISTINCT STORE,   
                    REFERRED ,
                    CUST_ID   
   FROM  @MyTable) p
            PIVOT ( COUNT (CUST_ID) FOR REFERRED in (WALKIN, RADIO)) as  pk
  ON pv.store = pk.STORE

--不使用 PIVOT

SELECT 
   STORE,
   SUM(CASE REFERRED WHEN  'RADIO' THEN 1 ELSE 0 END )AS RADIO,
   SUM(CASE REFERRED WHEN 'WALKIN' THEN 1 ELSE 0 END )AS WALKIN,
   SUM(CASE REFERRED WHEN  'RADIO' THEN MONEYMADE ELSE 0 END )AS RADIO_MONEYMADE,
   SUM(CASE REFERRED WHEN 'WALKIN' THEN MONEYMADE ELSE 0 END )AS WALKIN_MONEYMADE
FROM   
   (

SELECT  
    STORE,   
    CUST_ID, 
    REFERRED ,
    SUM(MONEYMADE )MONEYMADE
  FROM 
    @MyTable p
  GROUP BY 
  STORE,
  REFERRED,
  CUST_ID


) t
group by store

两种技术的输出

    STORE       RADIO       WALKIN      RADIO_MONEY           WALKIN_MONEY
    ----------- ----------- ----------- --------------------- ---------------------
    1           1           1           40.00                 20.00
    2           1           0           90.00                 0.00
    3           0           2           0.00                  130.00

Answer 2

轮换 SQL 结果至少很棘手。如果您想直接从查询中呈现数据，您可能不得不牺牲查询的灵活性并硬编码大量信息（如列名及其计数）。

所以我想建议的是在实际使用查询的后端做更多的事情。如果可以接受并且您可以自己轮换结果，那么您可以执行以下操作：

;with R(Referred) as (
  select distinct REFERRED from MyTable
)
select t.STORE, COUNT(*) as COUNT, t.REFERRED
from MyTable t
  left join R on r.REFERRED = t.REFERRED
group by t.STORE, t.REFERRED

Answer 3

我认为，您的脚本已修改为符合您想要实现的目标：

SELECT 
   STORE, 
   COUNT(DISTINCT CASE REFERRED WHEN 'RADIO' THEN CUS_ID END) AS RADIO,  
   SUM  (CASE REFERRED WHEN 'RADIO' THEN MONEYMADE ELSE 0 END) AS RADIO_MONEY,  
   COUNT(DISTINCT CASE REFERRED WHEN 'WALKIN' THEN CUS_ID END) AS WALKIN
   SUM  (CASE REFERRED WHEN 'WALKIN' THEN MONEYMADE ELSE 0 END) AS WALKIN_MONEY, 
FROM MyTable
GROUP BY STORE