Question

我有下表名为＆＃34; Animal＆＃34;。

Animal Drink
Dog    Water
Dog    Water
Dog    Water
Cat    Milk
Cat    Milk
Cat    Water

我正在尝试根据动物饮用的饮料类型创建类别，结果将是。

Animal Drink
Dog    Water
Cat    Mixed

我尝试了各种案例陈述，但都没有成功。问题是我正在比较行（而不是列），这对我来说已经证明是非常困难的。

基本上，我想挑出只喝水的动物。如果一只动物只喝水，那么它就会被归类为水＃34;水。如果动物喝水和其他任何东西，那么它将被归类为＆＃34;混合＆＃34;。如果动物不喝任何水，那么它将被分类为“不能喝水”＃34;

Answer 1

为此，您只需按表达式使用组：

SELECT
  animal,
  CASE
    WHEN (sum_other = 0) THEN 'Water'
    WHEN (sum_water = 0) THEN 'NonWater'
    ELSE 'Mixed'
  END
FROM (SELECT
  animal,
  SUM(CASE
    WHEN (drink = 'Water') THEN 1
    ELSE 0
  END) AS sum_water,
  SUM(CASE
    WHEN (drink <> 'Water') THEN 1
    ELSE 0
  END) AS sum_other
FROM my_table
GROUP BY animal) A

Answer 2

可能看起来很复杂，但我认为会工作，虽然我没有测试过：

SELECT
      A.[Animal]
    , [Drink] = IIF(A.[Cnt] > 1, 'Mixed', T.[Drink]) 
FROM
    (
    SELECT
          [Animal]
        , [Cnt] = Count(*)  
    FROM
        [Table]
    GROUP BY
          [Animal]
        , [Drink]
    ) AS A
    OUTER APPLY(
        SELECT TOP 1 
            *
        FROM
            [Table] AS T
        WHERE
            A.Animal = T.Animal
    ) AS T

Answer 3

按动物计算饮料数量并使用CASE：

SELECT Animal,
CASE WHEN COUNT(DISTINCT Drink)=1
THEN Drink ELSE 'Mixed' END AS Drink
FROM T
GROUP BY Animal

Answer 4

也许是这样的：


SELECT 
t.Animal
,'Mixed'
FROM #TEMP t
GROUP BY t.Animal HAVING count(DISTINCT t.Drink) > 1
UNION ALL
SELECT 
t.Animal
,(SELECT Top 1 t1.Drink FROM #TEMP t1 WHERE t1.Animal = t.Animal) 
FROM #TEMP t
GROUP BY t.Animal HAVING COUNT(DISTINCT t.Drink) = 1

Answer 5

or something like this

select animal, drink from 
(select animal,
    rank() over (partition by animal order by  drink desc) as rn,
      case when rank() over (partition by animal order by  drink ) = 1
      then drink
      else 'mixed'
      end as drink
 from animal
)x
where x.rn = 1
group by 1, 2

Answer 6

简易解决方案：

SELECT Animal,
CASE WHEN COUNT(DISTINCT Drink)=1
THEN 
(SELECT DISTINCT Drink from my_table where Animal = T.Animal)
 ELSE 'Mixed' END AS Drink
FROM my_table as T
GROUP BY Animal

基于SQL中的行创建类别

6 个答案: