Question

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，其中一列名为“_date”，类型为DATE。

使用什么查询可以用所有期望的日期填充表格（而不必手动输入）？

Answer 1

试试这个：

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');

以下是使用它的SQL小提琴：http://sqlfiddle.com/#!2/65d13/1

按Andrew Fox的要求编辑（以检查日期是否已存在）。

CREATE PROCEDURE filldates(dateStart DATE, dateEnd DATE)

BEGIN

DECLARE adate date;

    WHILE dateStart <= dateEnd DO

        SET adate = (SELECT mydate FROM MyDates WHERE mydate = dateStart);

        IF adate IS NULL THEN BEGIN

            INSERT INTO MyDates (mydate) VALUES (dateStart);

        END; END IF;

        SET dateStart = date_add(dateStart, INTERVAL 1 DAY);

    END WHILE;

END;//

以下是使用它的SQL小提琴：http://sqlfiddle.com/#!2/66f86/1

Answer 2

我不希望我的SQL查询需要外部依赖（需要有一个日历表，用日期填充临时表的程序等）。这个查询的最初想法来自http://jeffgarretson.wordpress.com/2012/05/04/generating-a-range-of-dates-in-mysql/我有略微优化了清晰度和易用性。

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

优化和扩展此表以用于其他用途很简单。如果您只需要一周的数据，您可以轻松摆脱数十和数百个表。

如果您需要更多数字，可以轻松添加数千个表格。您只需要复制并粘贴数百个表，并添加零到9个数字。

Answer 3

我发现这个粘贴式变体工作：

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');

Answer 4

如果你遇到像我这样禁止程序的情况，并且你的sql用户没有插入权限，那么插入不允许，但是你想生成一个特定时期的日期列表，比如当年做一些聚合，使用这个

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'

Answer 5

如果你有一个足够大的连续id组的表，你可以使用 -

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，这可能会让你在闰年（366天）遇到麻烦

Answer 6

感谢我和我。我有一个更好的解决方案，允许您创建指定的日历表。例如，如果我想创建一个2014-04的表，它看起来像这样：

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')

Answer 7

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在变量中，并通过在每个循环上添加+1天让循环遍历它，例如，您将从2011年1月1日开始，然后循环将第一次加0，接下来加1天，接着是2天，依此类推$ i变量。然后，您可以打印出日期或将它们添加到数据库中。在这种情况下，$ i代表计数器，0表示起始点，＆lt; = 365表示您希望通过的循环数等于或小于天数，$ i ++为$ i添加+1每个循环都有变量。

日期（'Ymd'将日期转换为yyyy-mm-dd。使用大写Y给你一个完整的4位数年份，而使用小写y将给你一年中的最后2位数。你想保留它按此顺序将其添加到mySQL的日期字段中。

strtotime（$ originalDate将日期解析为Unix时间戳，而。“+”。$ i。“day”）基本上将$ i的值添加到日期中。

最后是mysqli查询。 $ db表示数据库连接变量，需要将其更改为您为连接设置的任何变量。接下来是实际查询。只需将您的表名称的单词表和VALUES之前的日期交换到日期行名称，就可以了。

以下是一个例子：

<?php
for($i=0;$i<=365;$i++){ 
$originalDate = "01/01/2011";
$date = date('Y-m-d',strtotime($originalDate . "+".$i." day"));
mysqli_query($db, "INSERT INTO table (date)VALUES('$date')");
}

使用for函数实现此目的的另一种方法是将strtotime日期直接包含在for动作中，作为与计数器变量相对的，这是一段更短的代码。将$ i = 0（起始计数点）替换为起始日期点，然后使用小于或等于结束日期点（循环次数），然后最后使用加号+1将第一个语句放入a变量可以使用。

最后，将日期转换为Y-m-d格式，准备放入数据库并运行查询。

同样，与第一个例子一样，这可以打印或直接放入数据库。

以下是一个例子：

<?php
for ($startdate = strtotime("2011-01-01"); $startdate <= strtotime("2011-12-31"); $startdate = strtotime("+1 day", $startdate)) {
$date= date("Y-m-d", $startdate);
mysqli_query($db, "INSERT INTO tracking (date)VALUES('$date')");
}

我可能觉得它听起来比实际上更令人困惑，但希望它至少会让你知道它是如何工作的。

Answer 8

我最近需要创建一个calendar_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，我使用以下查询在January 1, 2001和December 31, 2100（包括两者）之间填写所有可能的日期：

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数据库中，INSERT查询只需几秒钟。希望这有助于某人。

Answer 9

受到IvanD很多人的启发，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')

Answer 10

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');

如何使用一系列日期填充表格？

10 个答案: