我有以下查询返回所选用户的逗号分隔的电子邮件ID列表。
CREATE TABLE TEST
(
ID INT PRIMARY KEY,
FIRSTNAME VARCHAR(30) NOT NULL,
LASTNAME VARCHAR(30) NOT NULL,
EMAIL VARCHAR(50)
)
INSERT INTO TEST VALUES (1, 'Tom', 'Lew', 'tom@hotmail.com')
INSERT INTO TEST VALUES (2, 'Tom', 'Lew', 'tom@hotmail.com')
INSERT INTO TEST VALUES (3, 'Jack', 'Stan', 'jstan@hotmail.com')
INSERT INTO TEST VALUES (4, 'Tom', 'Reed', 'tomreed@hotmail.com')
INSERT INTO TEST VALUES (5, 'Tom', 'Reed', 'tomr@hotmail.com')
INSERT INTO TEST VALUES (6, 'Tom', 'Reed', 'treed@hotmail.com')
DECLARE @result VARCHAR(MAX)
SELECT @result = COALESCE(@result + ',', '') + [EMAIL]
FROM [TEST]
WHERE [FirstName] = 'Tom' AND [LastName] = 'Reed'
SELECT @result
我需要修改查询,以便它只返回一个值 - 如果只有一个与“Tom Reed”相关联的电子邮件ID,则会显示一封电子邮件ID - 如果为'Tom Reed'找到多个电子邮件ID,则为字符串'multiple'
我以某种方式能够获得理想的结果但不确定性能。在实际情况中,此表将包含数千条记录,并将连接到其他表。
SELECT
DISTINCT
CASE
WHEN CNT = 1 THEN EMAIL
ELSE 'MULTIPLE'
END
FROM (SELECT
ID, EMAIL,
COUNT(*) OVER() CNT
FROM TEST
WHERE [FirstName] = 'Tom' AND [LastName] = 'Reed') T
很抱歉,如果主题行和内容不匹配。不确定描述。
期望的结果:
如果查询'Tom Reed'
Output: Multiple
如果查询'Tom Lew'
Output: Multiple
如果查询'杰克斯坦'
Output: jstan@hotmail.com
答案 0 :(得分:2)
我认为这可行(未经测试):
SELECT DISTINCT
FirstName,
LastName,
CASE
WHEN COUNT(Email) OVER (PARTITION BY FirstName,LastName) = 1
THEN Email
ELSE 'Multiple' END
FROM Test
如果你只需要传递参数的单词:
SELECT DISTINCT
CASE
WHEN COUNT(Email) OVER (PARTITION BY FirstName,LastName) = 1
THEN Email
ELSE 'Multiple' END
FROM Test
WHERE FirstName = 'Tom'
AND LastName = 'Reed'
这应该返回电子邮件或单词Multiple。