如何使用db2中的group by子句做COUNT（*）？

Question

我想知道有多少行如何去这个查询返回：

SELECT 
    MONTH(DHSTMP), SUM(DHDLDY), SUM(DHBUDS), YEAR(DHSTMP) 
FROM 
    DSHDAY 
WHERE 
    DHSTMP BETWEEN '2009-07-26 00:00:00' AND '2011-03-09 23:59:59' 
GROUP BY 
    YEAR(DHSTMP), MONTH(DHSTMP) ORDER BY YEAR(DHSTMP), MONTH(DHSTMP) 

所以，我想：

SELECT 
    COUNT(*) 
FROM 
    DSHDAY 
WHERE 
    DHSTMP BETWEEN '2009-07-26 00:00:00' AND '2011-03-09 23:59:59' 
GROUP BY 
    YEAR(DHSTMP), MONTH(DHSTMP) 

但它告诉我多少行每个组。

如何获取第一个查询将返回的行数？

Answer 1

这应该工作。但它不是高性能的。
（如果你只在计数兴趣，你可能不希望计算和汇总和排序）

SELECT COUNT(*) 
FROM (
    SELECT 
    MONTH(DHSTMP), 
    SUM(DHDLDY), 
    SUM(DHBUDS), 
    YEAR(DHSTMP) 
    FROM DSHDAY 
    WHERE DHSTMP BETWEEN '2009-07-26 00:00:00' AND '2011-03-09 23:59:59' 
    GROUP BY 
    YEAR(DHSTMP), 
    MONTH(DHSTMP) 
    ORDER BY 
    YEAR(DHSTMP), 
    MONTH(DHSTMP) 
) i 

试试这个

SELECT COUNT(DISTINCT i.month_year) 
FROM (
    SELECT 
    CONCAT(MONTH(DHSTMP), YEAR(DHSTMP)) AS month_year 
    FROM DSHDAY 
    WHERE DHSTMP BETWEEN '2009-07-26 00:00:00' AND '2011-03-09 23:59:59' 
) i 

变化（在回答迈克的评论）

SELECT COUNT(*) 
FROM (
    SELECT 0 
    FROM DSHDAY 
    WHERE DHSTMP BETWEEN '2009-07-26 00:00:00' AND '2011-03-09 23:59:59' 
    GROUP BY 
    YEAR(DHSTMP), 
    MONTH(DHSTMP) 
) i