Mysql从日期范围列表中生成每个日期答案

【问题标题】：Mysql Generate each date from date range listMysql从日期范围列表中生成每个日期
【发布时间】：2016-04-18 20:58:30
【问题描述】：

我有一个查询（从 bla.. bla.. 中选择 *）产生这样的日期范围结果：

code | date1 | date2

a | 2016-04-19  | 2016-04-21 |

b | 2016-04-13  | 2016-04-14 |

我想在 date1 和 date2 之间生成该日期范围的每一天，如下所示：

代码 |日期结果

a | 2016-04-19

a | 2016-04-20

a | 2016-04-21

b | 2016-04-13

b | 2016-04-14

我找到了在两个日期范围之间产生每个日期的查询示例，如下所示：

SELECT ADDDATE('2016-04-10', INTERVAL @i:=@i+1 DAY) AS DAY
FROM (
SELECT a.a
FROM (SELECT 0 AS a 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) AS a
CROSS JOIN (SELECT 0 AS a 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) AS b
CROSS JOIN (SELECT 0 AS a 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) AS c
) a
JOIN (SELECT @i := -1) r1
WHERE 
@i < DATEDIFF('2016-04-19', '2016-04-10')

但我无法用我的查询来实现它:(

【问题讨论】：

标签： mysql date

【解决方案1】：

您可以使用 from_days() 将日期转换为天数然后使用计数表进行内部连接（序号从 1 开始）数字 730485 是 '2000-01-01' 偏移量（select from_days('2000-01-01')）

select a.* , from_days(t.tallyid+730485) from 
(
    select 'a' code , '2016-04-19' date1,  '2016-04-21' date2
    union all
    select 'b'code , '2016-04-13' date1,  '2016-04-14' date2
) a
inner join Tally t on t.tallyid between (TO_DAYS(a.date1)-730485) and (TO_DAYS(a.date2)-730485)

【讨论】：

很有趣.. 但很抱歉，如果我不想创建像计数表这样的临时表，我怎样才能让它在单个查询中工作？

【解决方案2】：

这里是单个查询：

select a.* , from_days(t.tallyid+730485) from 
(
    select 'a' code , '2016-04-19' date1,  '2016-04-21' date2
    union all
    select 'b'code , '2016-04-13' date1,  '2016-04-14' date2,
) a
left join 
(
    SELECT @row := @row + 1 as tallyid FROM 
    (select 0 union all select 1 union all select 3 union all select 4 union all select 5 union all select 6 union all select 6 union all select 7 union all select 8 union all select 9) t,
    (select 0 union all select 1 union all select 3 union all select 4 union all select 5 union all select 6 union all select 6 union all select 7 union all select 8 union all select 9) t2, 
    (select 0 union all select 1 union all select 3 union all select 4 union all select 5 union all select 6 union all select 6 union all select 7 union all select 8 union all select 9) t3, 
    (select 0 union all select 1 union all select 3 union all select 4 union all select 5 union all select 6 union all select 6 union all select 7 union all select 8 union all select 9) t4, 
    (select 0 union all select 1 union all select 3 union all select 4 union all select 5 union all select 6 union all select 6 union all select 7 union all select 8 union all select 9) t5, 
    (SELECT @row:=0) a
) t on t.tallyid between (TO_DAYS(a.date1)-730485) and (TO_DAYS(a.date2)-730485)

从 1 到 100000 的序列是在子查询中创建的，它适用于 2000 年到 2237 年之间的日期。

【讨论】：

【解决方案3】：

最后我找到了生成从 date1 到 date2 的日期的答案：

select 
*
from (
    select t.*, t.date1 + INTERVAL (a.a + (10 * b.a) + (100 * c.a)) DAY as tanggal
    from (select 0 as a 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) as a
    cross join (select 0 as a 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) as b
    cross join (select 0 as a 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) as c
    join (
        select date1,date2 from mytable
    ) t
    where t.date1 + INTERVAL (a.a + (10 * b.a) + (100 * c.a)) DAY <= t.date2
) a

谢谢你们的赞赏

【讨论】：