从日期范围生成天数

Question

我想运行一个类似的查询

select ... as days where `date` is between '2010-01-20' and '2010-01-24'

并返回如下数据:

days
----------
2010-01-20
2010-01-21
2010-01-22
2010-01-23
2010-01-24

Answer 1

此解决方案不使用循环,过程或临时表.子查询生成最近10,000天的日期,并且可以根据需要延伸到后退或前进.

select a.Date 
from (
    select curdate() - INTERVAL (a.a + (10 * b.a) + (100 * c.a) + (1000 * d.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
    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 d
) a
where a.Date between '2010-01-20' and '2010-01-24' 

输出:

Date
----------
2010-01-24
2010-01-23
2010-01-22
2010-01-21
2010-01-20

关于绩效的说明

在这里测试,性能出奇的好:上面的查询需要0.0009秒.

如果我们扩展子查询以生成约.100,000个数字(因此大约274年的日期),它运行0.0458秒.

顺便提一下,这是一种非常便携的技术,可以对大多数数据库进行微调.

返回1,000天的SQL小提琴示例

Answer 2

以下是使用视图的另一种变体:

CREATE VIEW digits AS
  SELECT 0 AS digit 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;

CREATE VIEW numbers AS
  SELECT
    ones.digit + tens.digit * 10 + hundreds.digit * 100 + thousands.digit * 1000 AS number
  FROM
    digits as ones,
    digits as tens,
    digits as hundreds,
    digits as thousands;

CREATE VIEW dates AS
  SELECT
    SUBDATE(CURRENT_DATE(), number) AS date
  FROM
    numbers;

然后你可以简单地做(看看它有多优雅？):

SELECT
  date
FROM
  dates
WHERE
  date BETWEEN '2010-01-20' AND '2010-01-24'
ORDER BY
  date

更新

值得注意的是,您只能从当前日期开始生成过去的日期.如果要生成任何类型的日期范围(过去,将来和之间),则必须使用此视图:

CREATE VIEW dates AS
  SELECT
    SUBDATE(CURRENT_DATE(), number) AS date
  FROM
    numbers
  UNION ALL
  SELECT
    ADDDATE(CURRENT_DATE(), number + 1) AS date
  FROM
    numbers;

Answer 3

接受的答案对PostgreSQL不起作用(语法错误在"a"或附近).

在PostgreSQL中执行此操作的方法是使用generate_series函数,即:

SELECT day::date
FROM generate_series('2010-01-20', '2010-01-24', INTERVAL '1 day') day;

    day
------------
 2010-01-20
 2010-01-21
 2010-01-22
 2010-01-23
 2010-01-24
(5 rows)

Answer 4

使用递归公用表表达式(CTE),您可以生成日期列表,然后从中进行选择.显然你通常不想创建三百万个日期,所以这只是说明了可能性.您可以简单地限制CTE内的日期范围,并使用CTE省略select语句中的where子句.

with [dates] as (
    select convert(datetime, '1753-01-01') as [date] --start
    union all
    select dateadd(day, 1, [date])
    from [dates]
    where [date] < '9999-12-31' --end
)
select [date]
from [dates]
where [date] between '2013-01-01' and '2013-12-31'
option (maxrecursion 0)

在Microsoft SQL Server 2005上,生成所有可能日期的CTE列表时间为1:08.生成一百年不到一秒钟.

Answer 5

MSSQL查询

select datetable.Date 
from (
    select DATEADD(day,-(a.a + (10 * b.a) + (100 * c.a)),getdate()) 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
) datetable
where datetable.Date between '2014-01-20' and '2014-01-24' 
order by datetable.Date DESC

产量

Date
-----
2014-01-23 12:35:25.250
2014-01-22 12:35:25.250
2014-01-21 12:35:25.250
2014-01-20 12:35:25.250

Answer 6

在没有循环/游标的情况下执行此操作的老式解决方案是创建一个NUMBERS表，该表具有单个 Integer 列，其值从 1 开始。

CREATE TABLE  `example`.`numbers` (
  `id` int(10) unsigned NOT NULL auto_increment,
  PRIMARY KEY  (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

您需要在表中填充足够的记录以满足您的需求：

INSERT INTO NUMBERS (id) VALUES (NULL);

获得NUMBERS表格后，您可以使用：

SELECT x.start_date + INTERVAL n.id-1 DAY
  FROM NUMBERS n
  JOIN (SELECT STR_TO_DATE('2010-01-20', '%Y-%m-%d') AS start_date 
          FROM DUAL) x
 WHERE x.start_date + INTERVAL n.id-1 DAY <= '2010-01-24'

绝对的低技术解决方案是：

SELECT STR_TO_DATE('2010-01-20', '%Y-%m-%d')
 FROM DUAL
UNION ALL
SELECT STR_TO_DATE('2010-01-21', '%Y-%m-%d')
 FROM DUAL
UNION ALL
SELECT STR_TO_DATE('2010-01-22', '%Y-%m-%d')
 FROM DUAL
UNION ALL
SELECT STR_TO_DATE('2010-01-23', '%Y-%m-%d')
 FROM DUAL
UNION ALL
SELECT STR_TO_DATE('2010-01-24', '%Y-%m-%d')
 FROM DUAL

你会用它做什么？

生成日期或数字列表以便左连接。您这样做是为了查看数据中存在间隙的位置，因为您左连接到序列数据列表 - 空值将使存在间隙的位置变得明显。