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：

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
当dateStart时，请尝试以下操作：

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，如果您的表中有足够大的连续ID集，那么在闰年（有366天）期间，这可能会给您带来麻烦-

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；
当crt_date时，我发现这种粘贴式变体可以工作：

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。您只需复制和粘贴带有数百个数字的表，并添加一个0到9的数字。
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）这个查询的最初想法是为了清晰和易于使用，我对其进行了一些优化

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')

多亏了伊万德。
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')

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表示以0为起点的计数器，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'

如果像我这样的情况下，过程被禁止，并且您的sql用户没有插入权限，因此，不允许插入，但您希望生成特定时期的日期列表，比如说当年进行聚合，请使用此

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。希望这对某人有所帮助。
我最近需要创建一个
日历\u日期
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`));

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');

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更好的了？回答你的问题，我可以使用php，但我真的不知道如何使用它
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');