Mysql 如何用一系列日期填充表格？
我需要一个MySQL表来保存2011-01-01和2011-12-31之间的所有日期。我创建了一个表，其中有一个列名为“_date”，键入date 我可以通过什么查询将所有所需的日期填充到表中（而不必手动输入）？尝试以下操作：Mysql 如何用一系列日期填充表格？,mysql,sql,date,range,populate,Mysql,Sql,Date,Range,Populate,我需要一个MySQL表来保存2011-01-01和2011-12-31之间的所有日期。我创建了一个表，其中有一个列名为“_date”，键入date 我可以通过什么查询将所有所需的日期填充到表中（而不必手动输入）？尝试以下操作： DROP PROCEDURE IF EXISTS filldates; DELIMITER | CREATE PROCEDURE filldates(dateStart DATE, dateEnd DATE) BEGIN WHILE dateStart <= d
DROP PROCEDURE IF EXISTS filldates;
DELIMITER |
CREATE PROCEDURE filldates(dateStart DATE, dateEnd DATE)
BEGIN
WHILE dateStart <= dateEnd DO
INSERT INTO tablename (_date) VALUES (dateStart);
SET dateStart = date_add(dateStart, INTERVAL 1 DAY);
END WHILE;
END;
|
DELIMITER ;
CALL filldates('2011-01-01','2011-12-31');
删除程序（如果存在）；
分隔符|
创建过程填充日期（日期开始日期、日期结束日期）
开始
而dateStart如果您有一个具有足够大的连续ID集的表，那么您可以使用它-
INSERT INTO tablename (_date)
SELECT '2011-01-01' + INTERVAL (id - 1) DAY
FROM some_table_with_lots_of_ids
WHERE id BETWEEN 1 AND 365
注意：但请注意，在闰年（有366天）期间，这可能会给您带来麻烦。我发现这种“粘贴并运行”变体有效：
DROP PROCEDURE IF EXISTS FillCalendar;
DROP TABLE IF EXISTS calendar;
CREATE TABLE IF NOT EXISTS calendar(calendar_date DATE NOT NULL PRIMARY KEY);
DELIMITER $$
CREATE PROCEDURE FillCalendar(start_date DATE, end_date DATE)
BEGIN
DECLARE crt_date DATE;
SET crt_date = start_date;
WHILE crt_date <= end_date DO
INSERT IGNORE INTO calendar VALUES(crt_date);
SET crt_date = ADDDATE(crt_date, INTERVAL 1 DAY);
END WHILE;
END$$
DELIMITER ;
CALL FillCalendar('2013-01-01', '2013-01-03');
CALL FillCalendar('2013-01-01', '2013-01-07');
删除程序（如果存在）；
如果存在日历，则删除表；
如果日历不存在，则创建表（日历\日期不为空主键）；
分隔符$$
创建过程填充日历（开始日期、结束日期）
开始
宣布crt_日期；
设置crt\u日期=开始日期；
虽然crt_date我不希望我的SQL查询需要外部依赖项（需要有一个日历表、用日期填充临时表的过程等），但这个查询的最初想法是为了清晰和易于使用而进行了一些优化
SELECT (CURDATE() - INTERVAL c.number DAY) AS date
FROM (SELECT singles + tens + hundreds number FROM
( SELECT 0 singles
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
) singles JOIN
(SELECT 0 tens
UNION ALL SELECT 10 UNION ALL SELECT 20 UNION ALL SELECT 30
UNION ALL SELECT 40 UNION ALL SELECT 50 UNION ALL SELECT 60
UNION ALL SELECT 70 UNION ALL SELECT 80 UNION ALL SELECT 90
) tens JOIN
(SELECT 0 hundreds
UNION ALL SELECT 100 UNION ALL SELECT 200 UNION ALL SELECT 300
UNION ALL SELECT 400 UNION ALL SELECT 500 UNION ALL SELECT 600
UNION ALL SELECT 700 UNION ALL SELECT 800 UNION ALL SELECT 900
) hundreds
ORDER BY number DESC) c
WHERE c.number BETWEEN 0 and 364
为其他用途优化和缩放此表非常简单。如果只需要一周的数据，就可以轻松地摆脱数十个和数百个表
如果您需要一组更大的数字，可以很容易地添加一个千表。您只需复制并粘贴带有数百个数字的表格，并添加一个0到9的数字。多亏了IvanD。
我有一个更好的解决方案，它允许您创建一个指定的日历表。
例如，如果我试图创建一个2014-04年的表，它如下所示：
SELECT (CURDATE() - INTERVAL c.number DAY) AS DATE
FROM
(
SELECT singles + tens + hundreds number FROM
(
SELECT 0 singles
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
) singles JOIN
(
SELECT 0 tens
UNION ALL SELECT 10 UNION ALL SELECT 20 UNION ALL SELECT 30
UNION ALL SELECT 40 UNION ALL SELECT 50 UNION ALL SELECT 60
UNION ALL SELECT 70 UNION ALL SELECT 80 UNION ALL SELECT 90
) tens JOIN
(
SELECT 0 hundreds
UNION ALL SELECT 100 UNION ALL SELECT 200 UNION ALL SELECT 300
UNION ALL SELECT 400 UNION ALL SELECT 500 UNION ALL SELECT 600
UNION ALL SELECT 700 UNION ALL SELECT 800 UNION ALL SELECT 900
) hundreds
ORDER BY number DESC
) c
WHERE c.number BETWEEN
DAYOFYEAR(NOW()) - DAYOFYEAR('2014-04-01')- DAY(LAST_DAY('2014-04-01')) +1
AND
DAYOFYEAR(NOW()) - DAYOFYEAR('2014-04-01')
受IvanD大量加入的启发，我来到这里：
SELECT DATE_ADD('2015-10-21', INTERVAL c.number DAY) AS DATE
FROM
(
SELECT singles + tens + hundreds+thousands number FROM
(
SELECT 0 singles
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
) singles JOIN
(
SELECT 0 tens
UNION ALL SELECT 10 UNION ALL SELECT 20 UNION ALL SELECT 30
UNION ALL SELECT 40 UNION ALL SELECT 50 UNION ALL SELECT 60
UNION ALL SELECT 70 UNION ALL SELECT 80 UNION ALL SELECT 90
) tens JOIN
(
SELECT 0 hundreds
UNION ALL SELECT 100 UNION ALL SELECT 200 UNION ALL SELECT 300
UNION ALL SELECT 400 UNION ALL SELECT 500 UNION ALL SELECT 600
UNION ALL SELECT 700 UNION ALL SELECT 800 UNION ALL SELECT 900
) hundreds
JOIN
(
SELECT 0 thousands
UNION ALL SELECT 1000 UNION ALL SELECT 2000 UNION ALL SELECT 3000
UNION ALL SELECT 4000 UNION ALL SELECT 5000 UNION ALL SELECT 6000
UNION ALL SELECT 7000 UNION ALL SELECT 8000 UNION ALL SELECT 9000
) thousands
ORDER BY number DESC
) c
WHERE c.number BETWEEN
0
AND
DATEDIFF('2016-10-08', '2015-10-21')
这可以通过使用简单的for循环在PHP中实现。有几种方法可以做到这一点。一种方法是将原始日期放置在变量中，并通过在每个循环中添加+1天，让循环每天运行它，例如，您将从2011年1月1日开始，然后循环将第一次添加0，第二次添加1天，然后在$i变量中添加2天等等。然后，您可以打印日期或将其添加到数据库中。在本例中，$i表示计数器，0为起点，如果您与我一样处于禁止执行过程的情况，并且您的sql用户没有插入权限，因此不允许插入，但您希望生成特定期间的日期列表，假设今年要进行一些聚合，请使用
select * from
(select adddate('1970-01-01',t4*10000 + t3*1000 + t2*100 + t1*10 + t0) gen_date from
(select 0 t0 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) t0,
(select 0 t1 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) t1,
(select 0 t2 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) t2,
(select 0 t3 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) t3,
(select 0 t4 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) t4) v
where gen_date between '2017-01-01' and '2017-12-31'
我最近需要创建一个calendar\u date
表，如下所示：
CREATE TABLE `calendar_date` (
`date` DATE NOT NULL -- A calendar date.
, `day` SMALLINT NOT NULL -- The day of the year for the date, 1-366.
, `month` TINYINT NOT NULL -- The month number, 1-12.
, `year` SMALLINT NOT NULL -- The year.
, PRIMARY KEY (`id`));
然后，我使用下面的查询将2001年1月1日至2100年12月31日（包括这两天）之间的所有可能日期填入其中：
INSERT INTO `calendar_date` (`date`
, `day`
, `month`
, `year`)
SELECT
DATE
, INCREMENT + 1
, MONTH(DATE)
, YEAR(DATE)
FROM
-- Generate all possible dates for every year from 2001 to 2100.
(SELECT
DATE_ADD(CONCAT(YEAR, '-01-01'), INTERVAL INCREMENT DAY) DATE
, INCREMENT
FROM
(SELECT
(UNITS + TENS + HUNDREDS) INCREMENT
FROM
(SELECT 0 UNITS 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) UNITS
CROSS JOIN
(SELECT 0 TENS UNION
SELECT 10 UNION SELECT 20 UNION SELECT 30 UNION
SELECT 40 UNION SELECT 50 UNION SELECT 60 UNION
SELECT 70 UNION SELECT 80 UNION SELECT 90) TENS
CROSS JOIN
(SELECT 0 HUNDREDS UNION
SELECT 100 UNION SELECT 200 UNION SELECT 300 UNION
SELECT 400 UNION SELECT 500 UNION SELECT 600 UNION
SELECT 700 UNION SELECT 800 UNION SELECT 900) HUNDREDS
) INCREMENT
-- For every year from 2001 to 2100, find the number of days in the year.
, (SELECT
YEAR
, DAYOFYEAR(CONCAT(YEAR, '-12-31')) - DAYOFYEAR(CONCAT(YEAR, '-01-01')) + 1 DAYS
FROM
-- Generate years from 2001 to 2100.
(SELECT
(2000 + UNITS + TENS) YEAR
FROM
(SELECT 0 UNITS 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) UNITS
CROSS JOIN
(SELECT 0 TENS UNION
SELECT 10 UNION SELECT 20 UNION SELECT 30 UNION
SELECT 40 UNION SELECT 50 UNION SELECT 60 UNION
SELECT 70 UNION SELECT 80 UNION SELECT 90) TENS
) YEAR
WHERE
YEAR BETWEEN 2001 AND 2100
) YEAR
WHERE
INCREMENT BETWEEN 0 AND DAYS - 1
ORDER BY
YEAR
, INCREMENT) DATE;
在我的本地MySQL数据库上，INSERT
查询只花了几秒钟。希望这对其他人有所帮助。而您只使用了mysql
？没有什么比php更有效的了？为了回答你的问题，我可以访问php，但是（我）不知道如何在php中快速计算范围内的所有日期，并且（ii）我想MySQL可以做得更高效？@hjpotter92：如果你对这两个方面都了解，那么在php中就真的不容易了。而且在MySQL中执行效率更高。至少在我的服务器上不起作用<代码>alter routine命令拒绝用户“name”@“localhost”对例程“name.filldates”执行：如果存在删除过程filldates，则这次出现另一个错误：您的SQL语法有错误；检查与您的MySQL服务器版本对应的手册，了解在第1行“DELIMITER | CREATE PROCEDURE filldates（dateStart DATE，dateEnd DATE）BEGIN WHI”附近使用的正确语法：DELIMITER | CREATE PROCEDURE filldates（dateStart DATE，dateEnd DATE）BEGIN，而dateStart正在运行您自己的MySQL服务器吗？如果是这样，请尝试以root用户身份登录，甚至向当前用户帐户授予必要的权限。示例：GRANT ALL ON mydb.*给'someuser'@'somehost'这里有更多关于这方面的信息：对于那些遇到语法错误的人。如果通过MySQL Workbench连接到数据库，并尝试运行创建过程脚本，则会出现语法错误。要运行此脚本，请右键单击“存储过程”菜单并选择“创建存储过程”，然后粘贴创建过程脚本，删除脚本末尾的“/”，然后单击“应用”。这对我很管用。瓦纳给你双倍的分数！这是一个非常有趣的概念，在处理许多日期时有点慢，但实际上相当不错。也许比其他选择更快。非常聪明。如果有人问自己：由内部生成的c表选择0-9、10-90、100-900之间的和，这是一个很大的间隔，数字从0到999（总共1000个）。外部选择是选择从0到364（即365天）的数字列表，并选择一年到今天的所有日期（CURDATE）。如果有人想要一些pi：调用FillCalendar（'2013-01-01'，'2099-01-04'）；从日历中选择计数（*）；31415这不会返回2014-04年的表格，而是返回所选月份和当前年份中已过去天数的表格！虽然这个代码片段可以解决这个问题，但它确实有助于提高文章的质量。请记住，您将在将来回答读者的问题，这些人可能不知道您的代码建议的原因。还请尽量不要用解释性注释挤满您的代码，因为这会降低代码和解释的可读性！很抱歉，这是我第一次发帖子。我对答案进行了编辑，添加了描述。我还回复了另一位在另一篇帖子中只想在一年中增加周一到周五的人。我想你可以在我的个人资料中找到…@Leniel Macaferi，你能帮我一下这是cleane吗
INSERT INTO my_dates (\`_date\`) SELECT DATE_ADD('2011-01-01', INTERVAL @_tmp:=@_tmp+1 day) \`_date\`
FROM (SELECT @_tmp:=-1 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) a /\*10^1\*/
JOIN (SELECT 0 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) b /\*10^2\*/
JOIN (SELECT 0 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) c /\*10^3\*/
WHERE @_tmp+1 BETWEEN 0 AND DATEDIFF('2011-12-31', '2011-01-01');
CREATE TABLE `calendar_date` (
`date` DATE NOT NULL -- A calendar date.
, `day` SMALLINT NOT NULL -- The day of the year for the date, 1-366.
, `month` TINYINT NOT NULL -- The month number, 1-12.
, `year` SMALLINT NOT NULL -- The year.
, PRIMARY KEY (`id`));
INSERT INTO `calendar_date` (`date`
, `day`
, `month`
, `year`)
SELECT
DATE
, INCREMENT + 1
, MONTH(DATE)
, YEAR(DATE)
FROM
-- Generate all possible dates for every year from 2001 to 2100.
(SELECT
DATE_ADD(CONCAT(YEAR, '-01-01'), INTERVAL INCREMENT DAY) DATE
, INCREMENT
FROM
(SELECT
(UNITS + TENS + HUNDREDS) INCREMENT
FROM
(SELECT 0 UNITS 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) UNITS
CROSS JOIN
(SELECT 0 TENS UNION
SELECT 10 UNION SELECT 20 UNION SELECT 30 UNION
SELECT 40 UNION SELECT 50 UNION SELECT 60 UNION
SELECT 70 UNION SELECT 80 UNION SELECT 90) TENS
CROSS JOIN
(SELECT 0 HUNDREDS UNION
SELECT 100 UNION SELECT 200 UNION SELECT 300 UNION
SELECT 400 UNION SELECT 500 UNION SELECT 600 UNION
SELECT 700 UNION SELECT 800 UNION SELECT 900) HUNDREDS
) INCREMENT
-- For every year from 2001 to 2100, find the number of days in the year.
, (SELECT
YEAR
, DAYOFYEAR(CONCAT(YEAR, '-12-31')) - DAYOFYEAR(CONCAT(YEAR, '-01-01')) + 1 DAYS
FROM
-- Generate years from 2001 to 2100.
(SELECT
(2000 + UNITS + TENS) YEAR
FROM
(SELECT 0 UNITS 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) UNITS
CROSS JOIN
(SELECT 0 TENS UNION
SELECT 10 UNION SELECT 20 UNION SELECT 30 UNION
SELECT 40 UNION SELECT 50 UNION SELECT 60 UNION
SELECT 70 UNION SELECT 80 UNION SELECT 90) TENS
) YEAR
WHERE
YEAR BETWEEN 2001 AND 2100
) YEAR
WHERE
INCREMENT BETWEEN 0 AND DAYS - 1
ORDER BY
YEAR
, INCREMENT) DATE;
INSERT INTO my_dates (\`_date\`) SELECT DATE_ADD('2011-01-01', INTERVAL @_tmp:=@_tmp+1 day) \`_date\`
FROM (SELECT @_tmp:=-1 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) a /\*10^1\*/
JOIN (SELECT 0 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) b /\*10^2\*/
JOIN (SELECT 0 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) c /\*10^3\*/
WHERE @_tmp+1 BETWEEN 0 AND DATEDIFF('2011-12-31', '2011-01-01');