Question

我需要在sql中生成两个日期之间的天数，如下所示：

我正在使用此查询：

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

效果很好，但总会产生1000天。我怎样才能生成像365这样的其他天数？

Answer 1

您可以添加限制条款（和订单）

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 
order by 1 desc 
limit 365

Answer 2

或者，如果允许以更易读的方式创建临时表 - 并且可以非常容易地扩展到10,000个整数或更多：

CREATE TEMPORARY TABLE units(idx) 
ENGINE=MEMORY 
AS (
          SELECT 0
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
);

SELECT a.the_date 
FROM (
SELECT curdate() - INTERVAL (hundreds + tens + units) DAY AS the_date
FROM       (SELECT idx      AS units    FROM units) units
CROSS JOIN (SELECT idx*  10 AS tens     FROM units) tens
CROSS JOIN (SELECT idx* 100 AS hundreds FROM units) hundreds
WHERE hundreds + tens + units < 365
) AS a
ORDER BY 1
;

从日期范围sql生成天数

2 个答案: