147

我想运行一个查询

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

并返回如下数据： 

天
----------
2010-01-20
2010-01-21
2010-01-22
2010-01-23
2010-01-24
4

29 回答 29

343

此解决方案不使用循环、过程或临时表。子查询会生成最近 10,000 天的日期，并且可以根据需要扩展为向前或向后。

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

性能说明

在这里测试一下，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大约。100,000 个数字（因此大约有 274 年的日期），它在 0.0458 秒内运行。

顺便说一句，这是一种非常便携的技术，只需稍作调整即可与大多数数据库一起使用。

返回 1,000 天的 SQL Fiddle 示例

于 2010-01-28T20:38:26.773 回答
34

这是使用视图的另一个变体：

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。如果您想生成任何类型的日期范围（过去、未来和介于两者之间），则必须改用此视图：

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;
于 2012-06-21T16:36:53.370 回答
29

接受的答案不适用于 PostgreSQL（“a”处或附近的语法错误）。

在 PostgreSQL 中执行此操作的方式是使用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)
于 2015-07-05T11:36:25.807 回答
17

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。显然您通常不想创建 300 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 从 select 语句中省略 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 列表需要 1:08。产生一百年的时间不到一秒钟。

于 2013-08-14T00:41:32.413 回答
7

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

输出

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
于 2014-09-28T08:41:15.013 回答
4

在没有循环/光标的情况下执行此操作的老式解决方案是创建一个NUMBERS表，该表有一个 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;

您需要使用足够的记录填充表以满足您的需求：

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 LEFT JOIN 到。您将这样做是为了查看数据中存在差距的位置，因为您将左连接到顺序数据列表中 - 空值将使存在差距的位置变得明显。

于 2010-01-28T19:51:00.920 回答
4

对于 Access 2010 - 需要多个步骤；我遵循与上面发布的相同模式，但认为我可以帮助 Access 中的某个人。对我来说效果很好，我不必保留种子表。

创建一个名为 DUAL 的表（类似于 Oracle DUAL 表的工作方式）

  • ID（自动编号）
  • DummyColumn（文本）
  • 添加一行值 (1,"DummyRow")

创建一个名为“ZeroThru9Q”的查询；手动输入以下语法：

SELECT 0 AS a
FROM dual
UNION ALL
SELECT 1
FROM dual
UNION ALL
SELECT 2
FROM dual
UNION ALL
SELECT 3
FROM dual
UNION ALL
SELECT 4
FROM dual
UNION ALL
SELECT 5
FROM dual
UNION ALL
SELECT 6
FROM dual
UNION ALL
SELECT 7
FROM dual
UNION ALL
SELECT 8
FROM dual
UNION ALL
SELECT 9
FROM dual;

创建一个名为“TodayMinus1KQ”的查询（对于今天之前的日期）；手动输入以下语法：

SELECT date() - (a.a + (10 * b.a) + (100 * c.a)) AS MyDate
FROM
  (SELECT *
   FROM ZeroThru9Q) AS a,

  (SELECT *
   FROM ZeroThru9Q) AS b,

  (SELECT *
   FROM ZeroThru9Q) AS c

创建一个名为“TodayPlus1KQ”的查询（对于今天之后的日期）；手动输入以下语法：

SELECT date() + (a.a + (10 * b.a) + (100 * c.a)) AS MyDate
FROM
  (SELECT *
   FROM ZeroThru9Q) AS a,

  (SELECT *
   FROM ZeroThru9Q) AS b,

  (SELECT *
   FROM ZeroThru9Q) AS c;

创建一个名为“TodayPlusMinus1KQ”的联合查询（日期为 +/- 1000 天）：

SELECT MyDate
FROM TodayMinus1KQ
UNION
SELECT MyDate
FROM TodayPlus1KQ;

现在您可以使用查询：

SELECT MyDate
FROM TodayPlusMinus1KQ
WHERE MyDate BETWEEN #05/01/2014# and #05/30/2014#
于 2014-05-22T16:49:01.260 回答
3

程序+临时表：

DELIMITER $$

CREATE DEFINER=`root`@`localhost` PROCEDURE `days`(IN dateStart DATE, IN dateEnd DATE)
BEGIN

    CREATE TEMPORARY TABLE IF NOT EXISTS date_range (day DATE);

    WHILE dateStart <= dateEnd DO
      INSERT INTO date_range VALUES (dateStart);
      SET dateStart = DATE_ADD(dateStart, INTERVAL 1 DAY);
    END WHILE;

    SELECT * FROM date_range;
    DROP TEMPORARY TABLE IF EXISTS date_range;

END
于 2014-06-12T15:32:16.890 回答
3

thx Pentium10 - 你让我加入了 stackoverflow :) - 这是我对 msaccess 的移植 - 认为它适用于任何版本：

SELECT date_value
FROM (SELECT a.espr1+(10*b.espr1)+(100*c.espr1) AS integer_value,
dateadd("d",integer_value,dateserial([start_year], [start_month], [start_day])) as date_value
FROM (select * from 
    (
    select top 1 "0" as espr1 from MSysObjects
    union all
    select top 1 "1" as espr2 from MSysObjects
    union all
    select top 1 "2" as espr3 from MSysObjects
    union all
    select top 1 "3" as espr4 from MSysObjects
    union all
    select top 1 "4" as espr5 from MSysObjects
    union all
    select top 1 "5" as espr6 from MSysObjects
    union all
    select top 1 "6" as espr7 from MSysObjects
    union all
    select top 1 "7" as espr8 from MSysObjects
    union all
    select top 1 "8" as espr9 from MSysObjects
    union all
    select top 1 "9" as espr9 from MSysObjects
    ) as a,
    (
    select top 1 "0" as espr1 from MSysObjects
    union all
    select top 1 "1" as espr2 from MSysObjects
    union all
    select top 1 "2" as espr3 from MSysObjects
    union all
    select top 1 "3" as espr4 from MSysObjects
    union all
    select top 1 "4" as espr5 from MSysObjects
    union all
    select top 1 "5" as espr6 from MSysObjects
    union all
    select top 1 "6" as espr7 from MSysObjects
    union all
    select top 1 "7" as espr8 from MSysObjects
    union all
    select top 1 "8" as espr9 from MSysObjects
    union all
    select top 1 "9" as espr9 from MSysObjects
    ) as b,
    (
    select top 1 "0" as espr1 from MSysObjects
    union all
    select top 1 "1" as espr2 from MSysObjects
    union all
    select top 1 "2" as espr3 from MSysObjects
    union all
    select top 1 "3" as espr4 from MSysObjects
    union all
    select top 1 "4" as espr5 from MSysObjects
    union all
    select top 1 "5" as espr6 from MSysObjects
    union all
    select top 1 "6" as espr7 from MSysObjects
    union all
    select top 1 "7" as espr8 from MSysObjects
    union all
    select top 1 "8" as espr9 from MSysObjects
    union all
    select top 1 "9" as espr9 from MSysObjects
    ) as c   
)  as d) 
WHERE date_value 
between dateserial([start_year], [start_month], [start_day]) 
and dateserial([end_year], [end_month], [end_day]);

引用 MSysObjects 只是因为在 from 子句中访问需要一个表计数至少 1 条记录 - 任何具有至少 1 条记录的表都可以。

于 2014-06-26T17:05:22.443 回答
3

在 MariaDB >= 10.3 和 MySQL >= 8.0 中使用新的递归（通用表表达式）功能的优雅解决方案。

WITH RECURSIVE t as (
    select '2019-01-01' as dt
  UNION
    SELECT DATE_ADD(t.dt, INTERVAL 1 DAY) FROM t WHERE DATE_ADD(t.dt, INTERVAL 1 DAY) <= '2019-04-30'
)
select * FROM t;

以上返回“2019-01-01”和“2019-04-30”之间的日期表。它也相当快。在我的机器上返回价值 1000 年的日期（约 365,000 天）大约需要 400 毫秒。

于 2019-02-05T16:30:00.963 回答
2

正如已经给出的许多精彩答案中所述（或至少暗示），一旦您有一组数字可供使用，这个问题就很容易解决。

注意：以下是 T-SQL，但这只是我对这里和整个互联网上已经提到的一般概念的特殊实现。将代码转换为您选择的方言应该相对简单。

如何？考虑这个查询：

SELECT DATEADD(d, N, '0001-01-22')
FROM Numbers -- A table containing the numbers 0 through N
WHERE N <= 5;

以上产生的日期范围为 1/22/0001 - 1/27/0001，非常简单。上述查询中有 2 条关键信息： 的开始日期和0001-01-22的偏移量5。如果我们结合这两条信息，那么我们显然有我们的结束日期。因此，给定两个日期，生成一个范围可以像这样分解：

  • 找到两个给定日期（偏移量）之间的差异，很容易：

    -- Returns 125 SELECT ABS(DATEDIFF(d, '2014-08-22', '2014-12-25'))

    在此处使用ABS()可确保日期顺序无关紧要。

  • 生成一组有限的数字，也很容易：

    -- Returns the numbers 0-2 SELECT N = ROW_NUMBER() OVER(ORDER BY (SELECT NULL)) - 1 FROM(SELECT 'A' AS S UNION ALL SELECT 'A' UNION ALL SELECT 'A')

    请注意，我们实际上并不关心我们在FROM这里选择什么。我们只需要一个集合来计算其中的行数。我个人使用 TVF，有些使用 CTE，有些使用数字表，你明白了。我提倡使用您也理解的最高效的解决方案。

结合这两种方法将解决我们的问题：

DECLARE @date1 DATE = '9001-11-21';
DECLARE @date2 DATE = '9001-11-23';

SELECT D = DATEADD(d, N, @date1)
FROM (
    SELECT N = ROW_NUMBER() OVER(ORDER BY (SELECT NULL)) - 1
    FROM (SELECT 'A' AS S UNION ALL SELECT 'A' UNION ALL SELECT 'A') S
) Numbers
WHERE N <= ABS(DATEDIFF(d, @date1, @date2));

上面的例子是可怕的代码，但展示了一切是如何结合在一起的。

更多乐趣

我需要经常做这种事情，所以我将逻辑封装到两个 TVF 中。第一个生成一个数字范围，第二个使用此功能生成一个日期范围。数学是为了确保输入顺序无关紧要，因为我想使用GenerateRangeSmallInt.

以下函数需要大约 16 毫秒的 CPU 时间来返回 65536 个日期的最大范围。

CREATE FUNCTION dbo.GenerateRangeDate (   
    @date1 DATE,   
    @date2 DATE   
)   
RETURNS TABLE
WITH SCHEMABINDING   
AS   
RETURN (
    SELECT D = DATEADD(d, N + 32768, CASE WHEN @date1 <= @date2 THEN @date1 ELSE @date2 END)
    FROM dbo.GenerateRangeSmallInt(-32768, ABS(DATEDIFF(d, @date1, @date2)) - 32768)
);

GO

CREATE FUNCTION dbo.GenerateRangeSmallInt (
    @num1 SMALLINT = -32768
  , @num2 SMALLINT = 32767
)
RETURNS TABLE
WITH SCHEMABINDING
AS
RETURN (
    WITH Numbers(N) AS (
        SELECT N FROM(VALUES
            (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1) -- 16
          , (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1) -- 32
          , (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1) -- 48
          , (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1) -- 64
          , (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1) -- 80
          , (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1) -- 96
          , (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1) -- 112
          , (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1) -- 128
          , (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1) -- 144
          , (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1) -- 160
          , (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1) -- 176
          , (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1) -- 192
          , (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1) -- 208
          , (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1) -- 224
          , (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1) -- 240
          , (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1) -- 256
        ) V (N)
    )
    SELECT TOP(ABS(CAST(@num1 AS INT) - CAST(@num2 AS INT)) + 1)
           N = ROW_NUMBER() OVER(ORDER BY (SELECT NULL)) + CASE WHEN @num1 <= @num2 THEN @num1 ELSE @num2 END - 1
    FROM Numbers A
       , Numbers B
);
于 2014-08-23T02:21:34.413 回答
2

试试这个。

SELECT TO_DATE('20160210','yyyymmdd') - 1 + LEVEL AS start_day 
from DUAL
connect by level <= (TO_DATE('20160228','yyyymmdd') + 1) - TO_DATE('20160210','yyyymmdd') ;
于 2016-07-28T09:16:45.740 回答
2

对于希望将此作为保存视图的任何人（MySQL 不支持视图中的嵌套选择语句）：

create view zero_to_nine as
    select 0 as n 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 date_range as
    select curdate() - INTERVAL (a.n + (10 * b.n) + (100 * c.n)) DAY as date
    from zero_to_nine as a
    cross join zero_to_nine as b
    cross join zero_to_nine as c;

然后你可以做

select * from date_range

要得到

date
---
2017-06-06
2017-06-05
2017-06-04
2017-06-03
2017-06-02
...
于 2017-06-06T12:48:54.717 回答
2

你想得到一个日期范围。

在您的示例中，您希望获取“2010-01-20”和“2010-01-24”之间的日期

可能的解决方案：

 select date_add('2010-01-20', interval row day) from
 ( 
    SELECT @row := @row + 1 as row FROM 
    (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 6 union all select 7 union all select 8 union all select 9) t,
    (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 6 union all select 7 union all select 8 union all select 9) t2, 
    (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 6 union all select 7 union all select 8 union all select 9) t3, 
    (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 6 union all select 7 union all select 8 union all select 9) t4, 
    (SELECT @row:=-1) r
 ) sequence
 where date_add('2010-01-20', interval row day) <= '2010-01-24'

解释

MySQL 有一个date_add函数，所以

select date_add('2010-01-20', interval 1 day)

会给你

2010-01-21

datediff函数会让你经常知道你必须重复这个

select datediff('2010-01-24', '2010-01-20')

返回

 4

获取日期范围内的日期列表归结为创建整数序列，请参阅在 MySQL 中生成整数序列

这里最受欢迎的答案采用了与https://stackoverflow.com/a/2652051/1497139类似的方法作为基础：

SELECT @row := @row + 1 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:=0) r
limit 4

这将导致

row
1.0
2.0
3.0
4.0

这些行现在可用于从给定的开始日期创建日期列表。要包括开始日期，我们从第 -1 行开始；

select date_add('2010-01-20', interval row day) from
 ( 
    SELECT @row := @row + 1 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:=-1) r
 ) sequence
 where date_add('2010-01-20', interval row day) <= '2010-01-24'
于 2018-11-29T14:23:26.137 回答
1

如果您需要几天以上的时间，您需要一张桌子。

在mysql中创建一个日期范围

然后，

select from days.day, count(mytable.field) as fields from days left join mytable on day=date where date between x and y;
于 2010-01-29T13:33:03.960 回答
1

在两个日期字段之间生成日期

如果您了解 SQL CTE 查询，那么此解决方案将帮助您解决问题

这是示例

我们在一张桌子上有日期

表名：“testdate”</p> 

STARTDATE   ENDDATE
10/24/2012  10/24/2012
10/27/2012  10/29/2012
10/30/2012  10/30/2012

要求结果：

STARTDATE
10/24/2012
10/27/2012
10/28/2012
10/29/2012
10/30/2012

解决方案：

WITH CTE AS
  (SELECT DISTINCT convert(varchar(10),StartTime, 101) AS StartTime,
                   datediff(dd,StartTime, endTime) AS diff
   FROM dbo.testdate
   UNION ALL SELECT StartTime,
                    diff - 1 AS diff
   FROM CTE
   WHERE diff<> 0)
SELECT DISTINCT DateAdd(dd,diff, StartTime) AS StartTime
FROM CTE

说明：CTE递归查询说明

  • 查询第一部分：

    SELECT DISTINCT convert(varchar(10), StartTime, 101) AS StartTime, datediff(dd, StartTime, endTime) AS diff FROM dbo.testdate

    说明：第一列是“开始日期”，第二列是开始和结束日期的天数之差，它将被视为“差异”列

  • 查询的第二部分：

    UNION ALL SELECT StartTime, diff-1 AS diff FROM CTE WHERE diff<>0

    说明：Union all 将继承上述查询的结果，直到结果为空，所以“StartTime”结果是从生成的 CTE 查询继承的，并且从 diff 中，减少 - 1，所以它看起来像 3、2 和 1直到 0

例如

STARTDATE   DIFF
10/24/2012  0
10/27/2012  0
10/27/2012  1
10/27/2012  2
10/30/2012  0

结果规范

STARTDATE       Specification
10/24/2012  --> From Record 1
10/27/2012  --> From Record 2
10/27/2012  --> From Record 2
10/27/2012  --> From Record 2
10/30/2012  --> From Record 3

  • 第三部分查询

    SELECT DISTINCT DateAdd(dd,diff, StartTime) AS StartTime FROM CTE

    它将在“startdate”中添加日期“diff”，因此结果应如下所示

结果

STARTDATE
10/24/2012
10/27/2012
10/28/2012
10/29/2012
10/30/2012
于 2012-10-27T08:52:33.210 回答
1

比接受的答案短，同样的想法：

(SELECT TRIM('2016-01-05' + INTERVAL a + b DAY) date
FROM
(SELECT 0 a UNION SELECT 1 a UNION SELECT 2 UNION SELECT 3
UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7
UNION SELECT 8 UNION SELECT 9 ) d,
(SELECT 0 b UNION SELECT 10 UNION SELECT 20
UNION SELECT 30 UNION SELECT 40) m
WHERE '2016-01-05' + INTERVAL a + b DAY  <=  '2016-01-21')
于 2016-01-21T09:23:39.953 回答
1

动态生成这些日期是个好主意。但是，我觉得自己在相当大的范围内不太舒服，所以我最终得到了以下解决方案：

  1. 创建了一个表格“DatesNumbers”，它将保存用于日期计算的数字：
CREATE TABLE DatesNumbers (
    i MEDIUMINT NOT NULL,
    PRIMARY KEY (i)
)
COMMENT='Used by Dates view'
;
  1. 使用上述技术填充表格，数字从 -59999 到 40000。这个范围将为我提供从 59999 天（~164 年）到 40000 天（109 年）的日期：
INSERT INTO DatesNumbers
SELECT 
    a.i + (10 * b.i) + (100 * c.i) + (1000 * d.i) + (10000 * e.i) - 59999 AS i
FROM 
  (SELECT 0 AS i 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,
  (SELECT 0 AS i 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,
  (SELECT 0 AS i 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,
  (SELECT 0 AS i 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,
  (SELECT 0 AS i 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 e
;
  1. 创建了一个视图“日期”：
SELECT
      i,
      CURRENT_DATE() + INTERVAL i DAY AS Date
FROM
    DatesNumbers

就是这样。

  • (+) 易于阅读的查询
  • (+) 没有即时数字世代
  • (+) 给出过去和未来的日期，并且在这篇文章中没有考虑到这一点。
  • (+) “仅在过去”或“仅在未来”日期可以使用WHERE i < 0或WHERE i > 0(PK)过滤
  • (-) 使用“临时”表和视图
于 2017-08-30T04:35:05.533 回答
1

适用于 AWS MySQL 的更通用的答案。

select datetable.Date
from (
    select date_format(adddate(now(),-(a.a + (10 * b.a) + (100 * c.a))),'%Y-%m-%d') 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 now() - INTERVAL 14 Day and Now()
order by datetable.Date DESC
于 2019-11-15T23:36:06.120 回答
0

好吧..试试这个： http: //www.devshed.com/c/a/MySQL/Delving-Deeper-into-MySQL-50/
http://dev.mysql.com/doc/refman/5.0/en/ loop-statement.html
http://www.roseindia.net/sql/mysql-example/mysql-loop.shtml

使用它来生成一个临时表，然后在临时表上执行 select *。或者一次输出一个结果。
你说你想做的事情不能用 SELECT 语句来完成，但它可能对 MySQL 特定的事情是可行的。
再说一次，也许你需要游标：http ://dev.mysql.com/doc/refman/5.0/en/cursors.html

于 2010-01-28T19:38:31.210 回答
0

对于 Oracle，我的解决方案是：

select trunc(sysdate-dayincrement, 'DD') 
  from dual, (select level as dayincrement 
                from dual connect by level <= 30)

Sysdate 可以更改为特定日期，并且可以更改级别编号以提供更多日期。

于 2013-10-29T16:57:29.847 回答
0

如果您想要两个日期之间的日期列表：

create table #dates ([date] smalldatetime)
while @since < @to
begin
     insert into #dates(dateadd(day,1,@since))
     set @since = dateadd(day,1,@since)
end
select [date] from #dates

*这里的小提琴：http ://sqlfiddle.com/#!6/9eecb/3469

于 2015-04-08T17:50:55.193 回答
0 
set language  'SPANISH'
DECLARE @table table(fechaDesde datetime , fechaHasta datetime ) 
INSERT @table VALUES('20151231' , '20161231');
WITH x AS 
    (
        SELECT   DATEADD( m , 1 ,fechaDesde ) as fecha  FROM @table
        UNION ALL
        SELECT  DATEADD( m , 1 ,fecha )
        FROM @table t INNER JOIN x ON  DATEADD( m , 1 ,x.fecha ) <= t.fechaHasta
    )
SELECT LEFT( CONVERT( VARCHAR, fecha , 112 ) , 6 ) as Periodo_Id 
,DATEPART ( dd, DATEADD(dd,-(DAY(fecha)-1),fecha)) Num_Dia_Inicio
,DATEADD(dd,-(DAY(fecha)-1),fecha) Fecha_Inicio
,DATEPART ( mm , fecha ) Mes_Id
,DATEPART ( yy , fecha ) Anio
,DATEPART ( dd, DATEADD(dd,-(DAY(DATEADD(mm,1,fecha))),DATEADD(mm,1,fecha))) Num_Dia_Fin
,DATEADD(dd,-(DAY(DATEADD(mm,1,fecha))),DATEADD(mm,1,fecha)) ultimoDia
,datename(MONTH, fecha) mes
,'Q' + convert(varchar(10),  DATEPART(QUARTER, fecha)) Trimestre_Name
FROM x 
OPTION(MAXRECURSION 0)
于 2015-12-02T15:46:57.747 回答
0 
DELIMITER $$
CREATE PROCEDURE GenerateRangeDates(IN dateStart DATE, IN dateEnd DATE)
BEGIN

    CREATE TEMPORARY TABLE IF NOT EXISTS dates (day DATE);

    loopDate: LOOP
        INSERT INTO dates(day) VALUES (dateStart); 
        SET dateStart = DATE_ADD(dateStart, INTERVAL 1 DAY);

        IF dateStart <= dateEnd 
            THEN ITERATE loopDate;
            ELSE LEAVE loopDate;
        END IF;
    END LOOP loopDate;

    SELECT day FROM dates;
    DROP TEMPORARY TABLE IF EXISTS dates;

END 
$$

-- Call procedure
call GenerateRangeDates( 
        now() - INTERVAL 40 DAY,
        now()
    );
于 2016-04-23T00:05:16.237 回答
0

SQLite版 RedFilters 顶级解决方案

select d.Date
from (
    select 
    date(julianday('2010-01-20') + (a.a + (10 * b.a) + (100 * c.a))) 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
) d
where 
d.Date between '2010-01-20' and '2010-01-24' 
order by d.Date
于 2016-06-20T08:55:13.590 回答
0

改进了工作日加入自定义假日表 microsoft MSSQL 2012 for powerpivot 日期表 https://gist.github.com/josy1024/cb1487d66d9e0ccbd420bc4a23b6e90e

with [dates] as (
    select convert(datetime, '2016-01-01') as [date] --start
    union all
    select dateadd(day, 1, [date])
    from [dates]
    where [date] < '2018-01-01' --end
)
select [date]
, DATEPART (dw,[date]) as Wochentag
, (select holidayname from holidaytable 
where holidaytable.hdate = [date]) 
as Feiertag
from [dates]
where [date] between '2016-01-01' and '2016-31-12'
option (maxrecursion 0)
于 2016-07-14T16:39:10.157 回答
0

使用递归公用表表达式的 mysql 8.0.1 和 mariadb 10.2.2 的另一种解决方案：

with recursive dates as (
    select '2010-01-20' as date
    union all
    select date + interval 1 day from dates where date < '2010-01-24'
)
select * from dates;
于 2018-11-28T09:10:39.297 回答
0 
WITH
  Digits AS (SELECT 0 D UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9),
  Dates AS (SELECT adddate('1970-01-01',t4.d*10000 + t3.d*1000 + t2.d*100 + t1.d*10 +t0.d) AS date FROM Digits AS t0, Digits AS t1, Digits AS t2, Digits AS t3, Digits AS t4)
SELECT * FROM Dates WHERE date BETWEEN '2017-01-01' AND '2017-12-31'
于 2019-02-02T10:00:55.130 回答
0

还可以创建一个程序来创建与日期不同的时间图的日历表。 如果您想要每个季度的表格

例如

2019-01-22 08:45:00
2019-01-22 09:00:00
2019-01-22 09:15:00
2019-01-22 09:30:00
2019-01-22 09:45:00
2019-01-22 10:00:00

您可以使用

CREATE DEFINER=`root`@`localhost` PROCEDURE `generate_calendar_table`()
BEGIN

select unix_timestamp('2014-01-01 00:00:00') into @startts;
select unix_timestamp('2025-01-01 00:00:00') into @endts;

if ( @startts < @endts ) then

    DROP TEMPORARY TABLE IF EXISTS calendar_table_tmp;

    CREATE TEMPORARY TABLE calendar_table_tmp (ts int, dt datetime); 

    WHILE ( @startts < @endts)
        DO 
        SET @startts = @startts + 900;
        INSERT calendar_table_tmp VALUES (@startts, from_unixtime(@startts));
    END WHILE;

END if;

END

然后通过操纵

select ts, dt from calendar_table_tmp;

那也给你ts

'1548143100', '2019-01-22 08:45:00'
'1548144000', '2019-01-22 09:00:00'
'1548144900', '2019-01-22 09:15:00'
'1548145800', '2019-01-22 09:30:00'
'1548146700', '2019-01-22 09:45:00'
'1548147600', '2019-01-22 10:00:00'

从这里您可以开始添加其他信息，例如

select ts, dt, weekday(dt) as wd from calendar_table_tmp;

或使用create table 语句创建真实表

于 2019-02-19T20:53:01.347 回答