Question

我已经编写了一个存储过程来获取该日期的一周，它还会返回一周开始的日期以及周数和年份。

我知道周末＆＃39;函数，但这并没有给我一周开始的日期，而且我不知道在一周和一年中这样做的函数。

问题是：
如何获得“约会”日期？在一周的开始给出周数？将周开始作为日期索引传递的地方，0 =星期日，1 =星期一等。

我目前的功能并不总是有效，如果一周的第一天是星期一，那么星期日会进入下一周，而不是我想要的同一周结束。

Answer 1

我也在对此进行了一些挖掘。但是我偶然发现了一些同样有效的mysql代码。它基本上是根据星期几减去几天。也就是说，如果日期是星期三（4），则您知道该日期是1-4 = -3天前。

如何？

# with Sunday being the start of the week:
select convert(date_add(now(), interval(1-dayofweek(now())) day), date) as WeekStartDate
select convert(date_add(now(), interval(7-dayofweek(now())) day), date) as WeekEndDate

# with Monday being the start of the week:
select convert(date_add(now(), interval(2-dayofweek(now())) day), date) as WeekStartDate
select convert(date_add(now(), interval(8-dayofweek(now())) day), date) as WeekEndDate

信用： How do I get the first day of the week of a date in mysql?

Answer 2

使用Sequence引擎。您可以根据需要调整以下示例：

MariaDB [_]> SHOW ENGINES\G
.
.
.
*************************** 3. row ***************************
      Engine: SEQUENCE
     Support: YES
     Comment: Generated tables filled with sequential values
Transactions: YES
          XA: NO
  Savepoints: YES
.
.
.

MariaDB [_]> SET @`year` := 2016,
          ->     @`mode` := 1,
          ->     @`week` := 23;
Query OK, 0 rows affected (0.00 sec)

MariaDB [_]> SELECT
          ->   `der`.`date`,
          ->   `der`.`week`,
          ->   `der`.`year`
          -> FROM (
          ->   SELECT
          ->     `der`.`date`,
          ->     WEEK(`der`.`date`, @`mode`) `week`,
          ->     YEAR(`der`.`date`) `year`
          ->   FROM (
          ->     SELECT
          ->       DATE_ADD(CONCAT(@`year`, '-01-01'), INTERVAL `s`.`seq` DAY) `date`
          ->     FROM
          ->       seq_0_to_365 `s`
          ->   ) `der`
          -> ) `der`
          -> WHERE
          ->   `der`.`week` = @`week` AND
          ->   `der`.`year` = @`year`;
+------------+------+------+
| date       | week | year |
+------------+------+------+
| 2016-06-06 |   23 | 2016 |
| 2016-06-07 |   23 | 2016 |
| 2016-06-08 |   23 | 2016 |
| 2016-06-09 |   23 | 2016 |
| 2016-06-10 |   23 | 2016 |
| 2016-06-11 |   23 | 2016 |
| 2016-06-12 |   23 | 2016 |
+------------+------+------+
7 rows in set (0.01 sec)

Answer 3

解决了，我重写了存储过程：

    exitProc:BEGIN
    #--
    # Procedure:
    #   weekFromDate
    #
    # Parameters:
    #   vcCompKey, the key associated with the company
    #   dtDate, the date to translate
    #   dtOutSOW, returned start of week date
    #   siOutWeek, returned week number
    #   siOutYear, returned year
    #--
        DECLARE siDIY          SMALLINT;   #Day in year
        DECLARE siFDOW         SMALLINT;   #First day of week
        DECLARE siGoBack       SMALLINT;      #Flag used to check for last year
        DECLARE siRmonth       SMALLINT;   #Reference Month
        DECLARE siRyear        SMALLINT;   #Reference Year
        DECLARE dtSOY          DATE;       #Date of start of year
        DECLARE vcFMDOY        VARCHAR(12);#First month and day of year
        DECLARE vcFDOW         VARCHAR(12);#First day of the week
        DECLARE vcDYSOW        VARCHAR(80);#Days of week
    #Get the first day of the week for the specified company
        SET vcFDOW = vcGetParamValue(vcCompKey, 'Var:First day of week');

    IF (vcFDOW IS NULL) THEN
    #No entry found, abort! 
        LEAVE exitProc;
    END IF;
    #Get the first month and day of the year for the specified company
    SET vcFMDOY = vcGetParamValue(vcCompKey, 'Var:First day of year');  

    IF (vcFMDOY IS NULL) THEN
    #No entry found, abort! 
        LEAVE exitProc;
    END IF;
    #Set-up days of week
    SET vcDYSOW = 'Sunday,Monday,Tuesday,Wednesday,Thursday,Friday,Saturday';   
    #Get the first day of the week index base 1
    SET siFDOW = FIND_IN_SET(LOWER(vcFDOW), LOWER(vcDYSOW)) - 1;
    #Get the reference month and year   
    SET siRmonth = MONTH(dtDate);
    SET siRyear = YEAR(dtDate);
    SET dtSOY = DATE(CONCAT(siRyear, '/', vcFMDOY)); 
    #Calculate the start of week date
    SET dtOutSOW = DATE_SUB(dtDate, INTERVAL (DAYOFWEEK(dtDate) - siFDOW) DAY) + 1;
    #Calculate the day in year
    SET siDIY = DATEDIFF(dtOutSOW, dtSOY);
    #Do we need to go back to the end of the previous year? 
    SET siGoBack = YEAR(dtDate) - YEAR(dtOutSOW);

    IF siGoBack < 0 Or siDIY < 0 Or dtDate < dtOutSOW THEN
    #Yes
        IF YEAR(dtOutSOW) = YEAR(dtDate) THEN
            SET dtOutSOW = DATE_SUB(dtOutSOW, INTERVAL 7 DAY);
        END IF;
        SET dtSOY = DATE(CONCAT(YEAR(dtOutSOW), '/', vcFMDOY)); 
        SET siDIY = DATEDIFF(dtOutSOW, dtSOY);
    END IF;     
    #Calculate the week no. and year
    SET siOutWeek = (siDIY / 7) + 1;
    SET siOutYear = YEAR(dtOutSOW); 
END

此例程确实利用了我的数据库中的其他表，并允许公司具有不同的年份。

Answer 4

作为测试，我会找到本周的开始，首先注意：

mysql> SELECT NOW(), WEEK(NOW());
+---------------------+-------------+
| NOW()               | WEEK(NOW()) |
+---------------------+-------------+
| 2016-06-18 12:10:58 |          24 |
+---------------------+-------------+

然后这是功能的核心：

mysql> SELECT '2016-01-01'
             + INTERVAL 7*24
             - DAYOFWEEK('2016-01-01')
             + 1 DAY;
+----------------------------------------------------------------+
| '2016-01-01' + INTERVAL 7*24 - DAYOFWEEK('2016-01-01') + 1 DAY |
+----------------------------------------------------------------+
| 2016-06-12                                                     |
+----------------------------------------------------------------+

'2016-01-01'是今年的开始。
24是WEEK()号码 + 1 DAY是为了弥补一周的开始还需要做一些其他事情来处理一周开始一周的选择。

MariaDB星期开始日期＆amp;第1周至第52周

4 个答案: