48

我需要一个 MySQL 表来保存 2011 年 1 月 1 日到 2011 年 12 月 31 日之间的所有日期。我创建了一个表，其中一列名称为“_date”，键入 DATE。

使用什么查询可以在表格中填充所有所需的日期（而不必手动输入）？

4

10 回答 10

54

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');

这是使用它的 SQL Fiddle：http ://sqlfiddle.com/#!2/65d13/1

按照Andrew Fox的要求编辑（检查日期是否已经存在）。

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;//

这是使用它的 SQL Fiddle：http ://sqlfiddle.com/#!2/66f86/1

于 2012-04-12T21:28:39.523 回答
27

我不希望我的 SQL 查询需要外部依赖项（需要有一个日历表、使用日期填充临时表的过程等）。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/为了清晰和易于使用，我对其进行了轻微优化。

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

为其他用途优化和扩展此表很简单。如果您只需要一周的数据，您可以轻松摆脱数十和数百张表。

如果您需要更大的数字集，添加千张表很容易。您只需要复制并粘贴数百个表格并添加一个零到九个数字。

于 2014-07-08T03:24:28.883 回答
23

如果您处于像我这样的程序被禁止的情况，并且您的 sql 用户没有插入权限，因此不允许插入，但是您想生成特定时期内的日期列表，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'
于 2017-09-01T01:39:44.473 回答
11

我发现这个粘贴后变体有效：

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');
于 2013-10-25T11:00:25.010 回答
5

我最近需要创建一个calendar_date如下表：

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填充了所有可能的日期January 1, 2001和December 31, 2100（包括）之间的所有日期：

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 数据库上，INSERT查询只用了几秒钟。希望这可以帮助某人。

于 2018-01-31T09:40:01.280 回答
2

如果你有一个表有足够大的连续 id 集，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，这可能会让您在闰年（有 366 天）期间遇到麻烦

于 2012-04-12T22:45:24.723 回答
2

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一个变量中，并通过在每个循环上添加 +1 天让循环每天运行，例如，您将从 01/01/2011 开始，然后循环将第一次添加 0，下一次添加 1 天，然后是 2 天，依此类推，以此类推到 $i 变量。然后您可以打印出日期或将它们添加到您的数据库中。在这种情况下，$i 将代表计数器，以 0 为起点，<=365 是您想要通过的循环数等于或小于天数，并且 $i++ 将 +1 添加到 $i 变量在每个循环上。

date('Ymd' 将日期转换为 yyyy-mm-dd。使用大写 Y 为您提供完整的 4 位数年份，而使用小写 y 将为您提供年份的最后 2 位数。您希望将其保留在此为了将它添加到 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（以天为单位）添加到日期。

最后是 mysqli 查询。$db 代表数据库连接变量，这将需要更改为您为连接设置的任何变量。接下来是实际的查询。只需将单词 table 替换为您的表名和 VALUES 之前的日期到您的日期行名称，您就可以开始了。

下面是一个例子：

<?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 操作中，作为对计数器变量的反对，这是一段更短的代码。将 $i=0 （起始计数器点）替换为起始日点，然后使用小于或等于结束日点（循环数），然后最后将第一条语句的 +1 放入 a变量准备使用。

最后，将日期转换为 Ymd 格式，准备好放入数据库并运行查询。

同样，与第一个示例一样，可以将其打印或直接放入您的数据库中。

下面是一个例子：

<?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')");
}

我可能让它听起来比实际更令人困惑，但希望它至少能让你了解它是如何工作的。

于 2016-08-08T01:43:47.913 回答
1

感谢伊万德。我有一个更好的解决方案，它允许您创建指定的日历表。例如，如果我试图创建一个 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')
于 2014-11-19T01:50:59.197 回答
0

受到 IvanD 的大量加入的启发，我得出了这个结论：

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')
于 2016-07-19T19:31:37.140 回答
0 
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');
于 2018-06-20T18:16:56.377 回答