Sql 返回一个值，该值指示列具有多个值

如何编写SQL查询来计算具有相同值的列，并返回指示该列具有多个值的值。见下面的例子

我的尝试：

SELECT ID, 'X' AS TYPE 
    FROM SAMPLE_TABLE 
    WHERE TYPE = 'C'
    GROUP BY ID HAVING COUNT(*) > 1 
    UNION 
    SELECT * FROM SAMPLE_TABLE  
    GROUP BY ID HAVING COUNT(*) = 1;

+-----------+
| ID | TYPE |
+-----------+
|101 |  C   |
|103 |  A   |
+-----------+

+-----------+
| ID | TYPE |
+-----------+
|101 |  C   |
|104 |  X   |
|105 |  X   |
+-----------+

样本

+-----------+
| ID | TYPE |
+-----------+
|101 |  C   |
|102 |  A   |
|102 |  B   |
|103 |  A   |
|104 |  B   |
|104 |  C   |
|105 |  A   |
|105 |  C   |
+-----------+

我的结果：

SELECT ID, 'X' AS TYPE 
    FROM SAMPLE_TABLE 
    WHERE TYPE = 'C'
    GROUP BY ID HAVING COUNT(*) > 1 
    UNION 
    SELECT * FROM SAMPLE_TABLE  
    GROUP BY ID HAVING COUNT(*) = 1;

+-----------+
| ID | TYPE |
+-----------+
|101 |  C   |
|103 |  A   |
+-----------+

+-----------+
| ID | TYPE |
+-----------+
|101 |  C   |
|104 |  X   |
|105 |  X   |
+-----------+

期望值：

SELECT ID, 'X' AS TYPE 
    FROM SAMPLE_TABLE 
    WHERE TYPE = 'C'
    GROUP BY ID HAVING COUNT(*) > 1 
    UNION 
    SELECT * FROM SAMPLE_TABLE  
    GROUP BY ID HAVING COUNT(*) = 1;

+-----------+
| ID | TYPE |
+-----------+
|101 |  C   |
|103 |  A   |
+-----------+

+-----------+
| ID | TYPE |
+-----------+
|101 |  C   |
|104 |  X   |
|105 |  X   |
+-----------+

---

您可以使用聚合和

案例

：

select id,
       (case when min(type) = max(type) then min(type) else 'X' end) as type
from t
group by id;

---

您可以使用条件聚合：

select id,
       case when sum(type <> 'C') then 'X' else 'C' end type
from tablename
group by id
having sum(type = 'C') > 0

---

结果中的第102行和第103行发生了什么？查询没有返回102行和第103行的结果，返回了“A”，我想返回包含“C”的ID，如果ID包含“C”和其他类型，则返回为“X”，其中“X”表示此ID包含“C”和其他类型。@PLOW。如果

id

的所有行上的类型都相同，则返回类型。否则，它将返回

'X'

。这就是我对这个问题的解释。如果我们想按“B”或“C”搜索，那么我们可以得到101:C、102:X、104:X和105:X检查这个：