Question

我有一个表定义如下：

CREATE TABLE A_TEST  (
    "ID"   NUMBER NOT NULL ENABLE,
    "TYPE" VARCHAR2(20 BYTE),
)

ID不是主键。
TYPE可能为NULL
ID我可以有0个，1个或更多TYPE。

如果我想获得不同类型的ID，我会写这个查询：

SELECT distinct a1.ID FROM A_TEST a1
  JOIN A_TEST a2 on a2.ID = a1.ID and a2.type != a1.type

但是我怎样才能找到我定义了单一类型的ID？（Id只有一行，id只有很多行，只有一行有定义的类型）

Answer 1

以下查询将返回表中只有一个TYPE的所有ID。

SELECT ID FROM A_TEST
WHERE TYPE IS NOT NULL -- since TYPE column can be NULL for a given ID.
GROUP BY ID
HAVING COUNT(*) = 1

E.g。

CREATE TABLE A_TEST ( ID INT, TYPE INT)
INSERT INTO A_TEST VALUES(1, NULL)
INSERT INTO A_TEST VALUES(2, 0)
INSERT INTO A_TEST VALUES(3, 0)
INSERT INTO A_TEST VALUES(3, 1)
INSERT INTO A_TEST VALUES(4, 1)
INSERT INTO A_TEST VALUES(4, 1)
INSERT INTO A_TEST VALUES(5, 0)
INSERT INTO A_TEST VALUES(5, NULL)

SELECT 
    ID 
FROM 
    A_TEST
WHERE 
    TYPE IS NOT NULL -- since TYPE column can be NULL for a given ID.
GROUP BY ID
HAVING COUNT(ID) = 1

结果

ID
-----------
2
4
(2 row(s) affected)

Answer 2

此示例考虑了NULL值类型：

SELECT ID
FROM A_TEST
WHERE TYPE IS NOT NULL
GROUP BY ID    
HAVING COUNT(DISTINCT TYPE)=1

例如，请参阅SQL Fiddle

Answer 3

在澄清之后，您的要求似乎是：

SELECT ID
FROM A_TEST
GROUP BY ID    
HAVING COUNT(DISTINCT TYPE) = 1
    OR COUNT(*) = 1 ;

根据where子句的出现次数查找行

3 个答案: