MySQL:选择范围内的所有日期,包括零增长天
可视化用户群增长时,包括没有增长的日子至关重要。为了在 MySQL 中实现这一目标,我们采用了一种称为笛卡尔积的技术。
考虑以下查询:
SELECT DATE(datecreated), count(*) AS number FROM users WHERE DATE(datecreated) > '2009-06-21' AND DATE(datecreated) <= DATE(NOW()) GROUP BY DATE(datecreated) ORDER BY datecreated ASC
此查询返回至少有一个用户的几天的结果。为了包含零增长的日期,我们创建了一个由笛卡尔积生成的日期表:
select date_add('2003-01-01 00:00:00.000', INTERVAL n5.num*10000+n4.num*1000+n3.num*100+n2.num*10+n1.num DAY ) as date from (select 0 as num union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) n1, (select 0 as num union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) n2, (select 0 as num union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) n3, (select 0 as num union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) n4, (select 0 as num union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) n5
这会生成一个包含从“2003-01-01”到“now()”的连续日期的表',允许我们与用户表执行左连接:
select * from ( select date_add('2003-01-01 00:00:00.000', INTERVAL n5.num*10000+n4.num*1000+n3.num*100+n2.num*10+n1.num DAY ) as date from (select 0 as num union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) n1, (select 0 as num union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) n2, (select 0 as num union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) n3, (select 0 as num union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) n4, (select 0 as num union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) n5 ) a where date >'2011-01-02 00:00:00.000' and date < NOW() order by date
此查询确保零增长天数出现在最终结果中结果。
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