如何修复不一致的 mysql 结果？

Question

我的MySQL查询

(SELECT * FROM names WHERE Rate = 75 ORDER BY RAND() LIMIT 1) 
UNION 
(SELECT * FROM names WHERE Rate = 75 ORDER BY RAND() LIMIT 1) 
UNION 
(SELECT * FROM names  WHERE Rate = 35 ORDER BY RAND() LIMIT 1);

它应该给我这样的东西

+----+-----+------+
| id | Rate| num  |
+----+-----+------+
| id |  75 |  987 |
| id |  75 |  987 |
| id |  35 |  987 |
+- --+-----+------+

但有时它只给我 2 行，像这样

+----+-----+------+
| id | Rate| num  |
+----+-----+------+
| id |  75 |  987 |
| id |  35 |  987 |
+- --+-----+------+

Answer 1

注意UNION和UNION ALL

的区别

• UNION 将删除重复的行
• UNION ALL 保留所有行

Answer 2

UNION 删除重复行。

前 2 个查询可能 return 同一行，在这种情况下，其中一个只会出现在结果中。

如果您不介意结果中出现重复行，则更改为 UNION ALL:

(SELECT * FROM names WHERE Rate = 75 ORDER BY RAND() LIMIT 1) 
UNION ALL 
(SELECT * FROM names WHERE Rate = 75 ORDER BY RAND() LIMIT 1) 
UNION ALL
(SELECT * FROM names WHERE Rate = 35 ORDER BY RAND() LIMIT 1);

如果您想要唯一行，则使用 LIMIT 2 将前 2 个查询统一为仅 1 个，这样它们 return 2 行：

(SELECT * FROM names WHERE Rate = 75 ORDER BY RAND() LIMIT 2) 
UNION ALL 
(SELECT * FROM names WHERE Rate = 35 ORDER BY RAND() LIMIT 1);

我在这里使用UNION ALL因为它比UNION.

性能更好

Answer 3

要从 table 中获取 3 不同的 行，只需执行以下操作：

SELECT * FROM names WHERE Rate = 35 ORDER BY RAND() LIMIT 3;