MySQL COUNT(*)GROUP BY COUNT =?

时间:2012-07-01 04:07:50

标签: mysql count group-by having-clause

这是我的查询

SELECT COUNT(*) as total, toys, date FROM T1
 WHERE (date >= '2012-06-26'AND date < '2012-06-30') AND (Avail > '0')
UNION
SELECT COUNT(*) as total, toys, date FROM T2
 WHERE (date >= '2012-06-26'AND date < '2012-06-30') AND (Avail > '0')
UNION
SELECT COUNT(*) as total, toys, date FROM T3
 WHERE (date >= '2012-06-26'AND date < '2012-06-30') AND (Avail > '0')
GROUP BY RoomType
HAVING COUNT( total ) = 4

输出结果

count   Toys            date
3   Bibi        2012-06-26
4   Baba            2012-06-26

如何让MYSQL仅显示count = 4

的结果

2 个答案:

答案 0 :(得分:4)

SELECT * FROM (
    SELECT COUNT(*) as total, toys, date FROM T1
     WHERE (date >= '2012-06-26' AND date < '2012-06-30') AND (Avail > '0')
     GROUP BY RoomType
    UNION
    SELECT COUNT(*) as total, toys, date FROM T2
     WHERE (date >= '2012-06-26' AND date < '2012-06-30') AND (Avail > '0')
     GROUP BY RoomType
    UNION
    SELECT COUNT(*) as total, toys, date FROM T3
     WHERE (date >= '2012-06-26'AND date < '2012-06-30') AND (Avail > '0')
    GROUP BY RoomType) AS src
WHERE total = 4;

请注意,对于正确的数据填充,必须将所有列都放在GROUP BY子句中或作为聚合函数的参数。 MySQL功能允许您避免此限制,但它可能会导致意外结果。

答案 1 :(得分:4)

首先需要分组然后结合结果。

SELECT COUNT(*) as total, toys, date FROM T1
 WHERE (date >= '2012-06-26'AND date < '2012-06-30') AND (Avail > '0')
 GROUP BY RoomType HAVING COUNT( *) = 4
UNION
SELECT COUNT(*) as total, toys, date FROM T2
 WHERE (date >= '2012-06-26'AND date < '2012-06-30') AND (Avail > '0')
 GROUP BY RoomType HAVING COUNT( * ) = 4
UNION
SELECT COUNT(*) as total, toys, date FROM T3
 WHERE (date >= '2012-06-26'AND date < '2012-06-30') AND (Avail > '0')
 GROUP BY RoomType HAVING COUNT( * ) = 4
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