在另一列中具有不同值的 Id 列需要以特殊值显示答案

【问题标题】：Id Column having diffrent values in another column need show with specilal value在另一列中具有不同值的 Id 列需要以特殊值显示
【发布时间】：2016-05-15 00:24:53
【问题描述】：

我在这个表 Id 列中有一个表测试，FuelId 和 FuelDesc 列在那里。值如下表所示，基于我需要创建如下表的前三列，请帮助我获取。

ID   FuelID   FuelDesc 
100   01       Elec     
101   02       Gas      
102   02       Gas      
100   02       Gas     
101   01       Elec    
103   01       Elec

O/P:-

ID    Pamenttype       
100   Both        
101   Both       
102   Gas         
103   Elec

【问题讨论】：

你试过了吗？如果有请分享。
SELECT ID, PaymentType = CASE WHEN COUNT(DISTINCT FuelId) > 1 THEN 'Both' ELSE MAX(FuelDESC) END FROM test GROUP BY ID

标签： mysql sql sql-server sql-server-2008

【解决方案1】：

WITH fuelCount
AS
(
    SELECT ID, COUNT(FuelID) AS fuelCount 
    FROM Test GROUP BY ID
)
SELECT fc.ID, 
CASE 
    WHEN fc.fuelCount > 1 THEN 'BOTH'
    ELSE (SELECT t.FuelDesc FROM Test t WHERE t.ID = fc.ID)
END AS Pamenttype
FROM fuelCount fc

【讨论】：

【解决方案2】：

您可以将Both 定义为一条记录为FuelID = '01'，另一条记录为FuelID = '02'。假设 {ID, FuelID} 在表 Test 中是唯一的：

with BothPaymentTypes as (
    select t1.ID
    from Test t1
    join Test t2 on t1.ID = t2.ID
    where t1.FuelID = '01'
    and t2.FuelID = '02'
)
select ID, 'Both' as Pamenttype
from BothPaymentTypes
union all
select ID, FuelDesc
from Test
where ID not in (select ID from BothPaymentTypes)

[SQL Fiddle Demo]

或者，您可以使用 COUNT 和 HAVING 子句避免自联接 - 再次假设 {ID, FuelID} 是唯一的：

with BothPaymentTypes as (
    select ID
    from Test
    group by ID
    having count(*) = 2
)
select ID, 'Both' as Pamenttype
from BothPaymentTypes
union all
select ID, FuelDesc
from Test
where ID not in (select ID from BothPaymentTypes)

[SQL Fiddle Demo]

【讨论】：

【解决方案3】：

您可以在COUNT 和MAX 函数的帮助下使用CASE 表达式：

SELECT
    ID,
    PaymentType = CASE WHEN COUNT(FuelId) > 1 THEN 'Both' ELSE MAX(FuelDESC) END
FROM test
GROUP BY ID

【讨论】：

【解决方案4】：

SELECT ID, CASE WHEN rowCount = 1 THEN FuelDesc ELSE 'Both' END AS Pamenttype
FROM
(SELECT ID, MAX(FuelDesc) AS FuelDesc, COUNT(*) AS rowCount
 FROM mytable
 GROUP BY ID)

【讨论】：