Question

我需要将此查询从MySQL格式转换为SQLite。我正在努力，但我发现了一些困难。

在SQLite中，curdate（）和interval函数不存在。

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 '2010-01-20' and '2010-01-24'

Answer 1

这个查询实际上做的只是生成大量连续日期（前几天最多一千天）。

在SQLite 3.8.3或更高版本中，使用递归common table expression可以更轻松地完成此操作：

WITH RECURSIVE dates(d)
AS (VALUES('2010-01-20')
    UNION ALL
    SELECT date(d, '+1 day')
    FROM dates
    WHERE d < '2010-01-24')
SELECT d AS date FROM dates;

Answer 2

以下是基本语法：

select a.Date 
from (select date('now', '-'||(a.a + (10 * b.a) + (100 * c.a))|| ' days') 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
           ) 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
           ) 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
           ) c
    ) a;

我遗漏了where条款，因为这些日期过去超过1000天，所以他们不会选择任何东西。

将查询从MySql转换为Sqlite

2 个答案: