Question

有人可以解释我如何SELECT一个日期，并在7 DAYS条款中给定日期范围WHERE的时间间隔前进？

这是我的示例SQL：

SELECT DATE(*TIMESTAMP*) + (DATE_ADD(DATE(*TIMESTAMP*), INTERVAL 7 DAY)) as `interval`
WHERE digital_lead.received <= '2014-12-16 08:37:00' AND
digital_lead.received >= '2014-11-14 08:37:00'

所以预期的输出会显示为

interval 
2014-11-14
2014-11-21
2014-11-28
2014-12-05

SELECT语句在7 DAYS子句中的日期范围前进WHERE。

有什么建议吗？

Answer 1

在这个答案的帮助下：https://stackoverflow.com/a/2652051/2307520来自类似的问题，我能够做到这一点：

select date_add('2014-11-14 08:37:00', INTERVAL (row) DAY) FROM
(
SELECT @row := @row + 7 as row 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 @row:=-7) q
WHERE @row<=DATEDIFF('2014-12-16 08:37:00' , '2014-11-14 08:37:00')
) z

如何通过7天MYSQL选择和前进

1 个答案: