Question

请考虑以下MySQL表

|----+--------+----------|
| id | result |reference | 
|----+--------+----------|
|  1 |ok      |       33 |
|  2 |ok      |       46 | 
|  3 |ko      |       55 |  
|  4 |ko      |       55 |  
|  5 |ok      |       55 |  
|  6 |ko      |       47 | 
|  7 |ko      |       89 |  
|  8 |ok      |       91 | 
|  9 |ko      |       47 |  
+----+--------+----------+

我想选择result = ko和reference = 47的行，因为引用47在result = ok的任何行中都不存在。除此之外我只需要这一行，因为这种情况会发生两次（id = 6和id = 9），尽管它可能会发生一次。实际上在我要求的查询中，id = 89的行也应该出现，作为reference = 89，result = ko，并且没有其他行带有reference = 89和result = ok。非常感谢你提前！

Answer 1

您可以使用subquery with not exists：

SELECT DISTINCT result, reference 
FROM [dbo].[references] as x
WHERE NOT EXISTS 
(SELECT * FROM [dbo].[references] as y
 WHERE
 y.reference = x.reference
 and
 y.result = 'ok');

您也可以使用自我左联接：

SELECT DISTINCT x.result, x.reference 
FROM [dbo].[references] as x
LEFT JOIN
[dbo].[references] as y 
on x.reference = y.reference
and y.result = 'ok'
WHERE
y.reference IS NULL

Answer 2

使用条件，有效：

IF EXISTS (Select TOP 1 * from <thistable>
WHERE result='OK' AND reference=47)
SELECT Select TOP 1 * from <thistable>
WHERE result='OK' AND reference=47
ELSE
SELECT Select TOP 1 * from <thistable>
WHERE result='KO' AND reference=47
END

选择具有列值的行仅在另一列具有特定值时才存在

2 个答案: