从日期范围生成天数

时间：2010-01-28 19:21:45

标签： sql mysql datetime between

我想运行像

这样的查询 
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

29 个答案:

答案 0 :(得分：298)

此解决方案使用无循环，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'

<强>输出： 

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

有关效果的说明

测试here，效果出奇的好：以上查询需要0.0009秒。

如果我们扩展子查询以生成约。 100,000个数字（因此约为274年的日期），它的运行时间为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。

SQL Fiddle example returning 1,000 days

答案 1 :(得分：31)

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

<强>更新

值得注意的是，您只能从当前日期生成过去的日期。如果您想生成任何类型的日期范围（过去，将来和之间），您将不得不使用此视图：

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;

答案 2 :(得分：20)

接受的答案对PostgreSQL没有用（语法错误等于或接近＆＃34; a＆＃34;）。

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)

答案 3 :(得分：13)

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，你通常不想创造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。

答案 4 :(得分：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

答案 5 :(得分：4)

没有循环/游标的旧学校解决方案是创建一个NUMBERS表，它有一个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格以满足您的需求：

INSERT INTO NUMBERS (id) VALUES (NULL);

获得NUMBERS表后，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将是：

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。您可以这样做，以便查看数据中存在间隙的位置，因为您将LEFT JOIN加入到顺序数据列表中 - 空值将明显存在间隙。

答案 6 :(得分：4)

对于Access 2010 - 需要多个步骤;我遵循上面发布的相同模式，但我想我可以帮助Access中的某个人。对我来说很好，我没有必要保留一个种子日期表。

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（文字）
  • 添加一行值（1，＆＃34; DummyRow＆＃34;）

创建名为＆＃34; ZeroThru9Q＆＃34 ;;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＆＃34; TodayMinus1KQ＆＃34;的查询（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

创建名为＆＃34; TodayPlus1KQ＆＃34;的查询（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;

创建名为＆＃34; TodayPlusMinus1KQ＆＃34;的联合查询（日期+/- 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#

答案 7 :(得分：3)

奔腾10 - 你让我加入stackoverflow :) - 这是我移植到msaccess - 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]);

引用了MSysObjects只是因为访问需要一个表countin＆＃39;在from子句中至少有1条记录 - 任何至少有1条记录的表都可以。

答案 8 :(得分：2)

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

答案 9 :(得分：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') ;

答案 10 :(得分：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

以下是示例

我们在一个表中有日期

表名：“testdate”

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

<强>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

  • 查询的第一部分：

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

    说明：firstcolumn是“startdate”，第二列是start和end的差异 以天为单位的日期，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result为null， 因此，“StartTime”结果继承自生成的CTE查询，并且来自diff，减少-1，因此它看起来像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

  • 查询的第3部分

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

    它将在“startdate”中添加日“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

答案 11 :(得分：1)

正如已经给出的许多精彩答案所述（或至少已经提到过），一旦你有一组数字可以解决这个问题就很容易解决。

注意：以下是T-SQL，但它只是我在这里和整个互联网上提到的一般概念的特定实现。将代码转换为您选择的方言应该相对简单。

如何？请考虑以下问题：

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。上述查询中有两个关键信息：0001-01-22的开始日期和5的偏移。如果我们将这两条信息结合起来，那么我们显然有结束日期。因此，给定两个日期，生成范围可以这样分解：

  • 找出两个给定日期（偏移量）之间的差异，简单：

    -- 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，也很简单：

    -- 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这里选择FROM的内容。我们只需要一个可以使用的集合，以便计算其中的行数。我个人使用TVF，有些人使用CTE，有些人则使用数字表，你就明白了。我主张使用您也理解的最高性能的解决方案。

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));

以上示例是可怕的代码，但演示了所有内容是如何组合在一起的。

更有趣

我需要做很多这样的事情，所以我将逻辑封装到两个TVF中。第一个生成一系列数字，第二个使用此功能生成一系列日期。数学是确保输入顺序无关紧要，因为我想使用GenerateRangeSmallInt中可用的全部数字。

以下函数需要大约16ms的CPU时间才能返回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
);

答案 12 :(得分：1)

如果你需要超过几天，你需要一张桌子。

Create a date range in mysql

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;

答案 13 :(得分：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')

答案 14 :(得分：1)

您想要获取日期范围。

在您的示例中，您希望获取介于'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'

说明

MySQL具有date_add功能，

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

会给你

2010-01-21

datediff函数会让您知道经常需要重复此操作

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，请参见generate an integer sequence in MySQL

此处最受好评的答案采用了与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

这些行现在可以用于创建从给定开始日期开始的日期列表。为了包括开始日期，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'

答案 15 :(得分：1)

对于任何想要将其作为已保存视图的人（MySQL不支持视图中的嵌套select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

获取

date
---
2017-06-06
2017-06-05
2017-06-04
2017-06-03
2017-06-02
...

答案 16 :(得分：1)

在飞行中生成这些日期是个好主意。但是，我觉得自己很不习惯在相当大的范围内做到这一点，所以我最终得到了以下解决方案：

  1. 创建了一个表“DatesNumbers”，用于保存用于日期计算的数字：

    2. 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.这个范围将给我的日期从59999天（〜164年）落后到40000天（109年）：

      2. 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”：

        2. SELECT i , CURRENT_DATE() + INTERVAL i DAY AS Date FROM DatesNumbers

        就是这样。

        • （+）易于阅读的查询
        • （+）暂时没有数字代数
        • （+）给出过去和未来的日期，并且在this post中可以看到没有UNION。
        • （+）“仅过去”或“仅限将来”日期可以使用WHERE i < 0或WHERE i > 0（PK）过滤
        • （ - ）'临时'表＆amp;使用视图

答案 17 :(得分：0)

一个适用于AWS MySQL的更通用的答案。

def addVariables(teachers, subjects):
  # convert ['CN','CS','L','I','EF','M'] to [('CN', 'CS'), ('L', 'I'), ('EF', 'M')]
  subjectPairs = zip(subjects[0::2], subjects[1::2])
  return dict(zip(teachers, subjectPairs))

print(addVariables(['Billy', 'Jannet', 'Julia'], ['CN','CS','L','I','EF','M']))
# prints {'Billy': ('CN', 'CS'), 'Jannet': ('L', 'I'), 'Julia': ('EF', 'M')}

答案 18 :(得分：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)

可以将Sysdate更改为特定日期，并且可以更改级别编号以提供更多日期。

答案 19 :(得分：0)

还可以创建一个程序，以创建时间戳记与日期不同的日历表。 如果您希望每个季度都有一张表

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

进行操作 
select ts, dt from calendar_table_tmp;

这也给你带来了快乐

'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

答案 20 :(得分：0)

在MariaDB> = 10.3和MySQL> = 8.0中使用新的递归（公用表表达式）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;

上面的代码返回一个介于'2019-01-01'和'2019-04-30'之间的日期表。它也相当快。在我的计算机上，返回价值1000年的日期（约365,000天）大约需要400毫秒。

答案 21 :(得分：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'

答案 22 :(得分：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;

答案 23 :(得分：0)

通过工作日改进加入自定义假期表 microsoft MSSQL 2012 for powerpivot date table 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)

答案 24 :(得分：0)

SQLite 版本的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

答案 25 :(得分：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()
    );

答案 26 :(得分：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)

答案 27 :(得分：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，但它可能适用于特定于MySQL的东西。
然后，也许你需要游标：http://dev.mysql.com/doc/refman/5.0/en/cursors.html

答案 28 :(得分：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

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

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