MySQL COUNT(*) GROUP BY HAVING COUNT=?答案

【问题标题】：MySQL COUNT(*) GROUP BY HAVING COUNT=?MySQL COUNT(*) GROUP BY HAVING COUNT=?
【发布时间】：2012-07-01 04:07:49
【问题描述】：

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

【问题讨论】：

标签： mysql count group-by having-clause

【解决方案1】：

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，对于正确的数据分组，您必须在GROUP BY 子句中或作为聚合函数的参数包含所有列。这是 MySQL 的特性，可以让你避免这种限制，但它可能会导致你意想不到的结果。

【讨论】：

【解决方案2】：

您需要先分组，然后合并结果。

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

【讨论】：