select a.Date from ( select curdate() - INTERVAL (a.a + (10 * b.a) + (100 * c.a)) DAY as Date 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 where a.Date between '2017-11-10' and '2017-11-15'
MySQL如何查询给定时间段内的每一天的日期?
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