Question

我需要使用mysql查找两个日期之间的星期日数。我知道如何使用PHP来实现它但是我需要使用mysql来计算它。

Answer 1

SET @START_DATE = '2014-01-22';
SET @END_DATE = '2014-06-29';
SELECT 
    ROUND((
      (unix_timestamp(@END_DATE) - unix_timestamp(@START_DATE) ) /(24*60*60)
      -7+WEEKDAY(@START_DATE)-WEEKDAY(@END_DATE)
    )/7)

 + if(WEEKDAY(@START_DATE) <= 6, 1, 0)
 + if(WEEKDAY(@END_DATE) >= 6, 1, 0) as Sundays;

解决方案由两部分组成：

计算完整周的数量（ROUND部分），显然你将在7周内有7个星期天。
包含在非完整周部分（第一个或最后一个）中的天数

如果您需要多次使用它，可以将其包装到函数中：

DROP function IF EXISTS `count_weekdays`;
DELIMITER $$
CREATE FUNCTION `count_weekdays` (startDate date, endDate date, wd int)
RETURNS INTEGER
BEGIN
RETURN ROUND((
      (unix_timestamp(endDate) - unix_timestamp(startDate) ) /(24*60*60)
      -7+WEEKDAY(startDate)-WEEKDAY(endDate)
    )/7)

 + if(WEEKDAY(startDate) <= wd, 1, 0)
 + if(WEEKDAY(endDate) >= wd, 1, 0);
END$$

DELIMITER ;

然后你就可以像这样使用它：

SET @START_DATE = '2018-07-03';
SET @END_DATE = '2018-07-28';
select 
  count_weekdays(@START_DATE, @END_DATE, 6) as Sundays,

  count_weekdays(@START_DATE, @END_DATE, 5) 
+ count_weekdays(@START_DATE, @END_DATE, 6) as weekends;

其中6~工作日代表星期日，5~星期六。

Answer 2

http://sqlfiddle.com/#!2/d41d8/50695

也许有一天这个链接上没有代码所以我在这里发布了。

上部链接将显示当天的天数和星期日的日期。如果你想找到任何其他日子，最后将1改为星期一的2

select DATE_ADD('2012-12-15', INTERVAL ROW DAY) as Date,
row+1  as DayOfMonth from (
SELECT @row := @row + 1 as row FROM 
(select 0 union all select 1 union all select 3 union all select 4 union all select 5 union all select 6) t1,
(select 0 union all select 1 union all select 3 union all select 4 union all select 5 union all select 6) t2, 
(SELECT @row:=-1) t3 limit 31
) b
where 
DATE_ADD('2012-12-01', INTERVAL ROW DAY)
between '2012-12-01' and '2012-12-31'
and
DAYOFWEEK(DATE_ADD('2012-12-01', INTERVAL ROW DAY))=1

Answer 3

这是否有效......

 SELECT FLOOR(
         (DATEDIFF(
            '@enddate'
           ,'@startdate' + INTERVAL 
            LENGTH(SUBSTRING_INDEX('654321',WEEKDAY('@startdate'),1)) 
              DAY) + 1)/7) + 1 x;

Answer 4

    SELECT a.StartDate, a.EndDate,
    (DAY(EndDate - StartDate) / 7)   
    +  iif(DAY(EndDate - StartDate)%7 + DATEPART(DW, StartDate) > 8 , 1, 0)
    Sundays FROM
    (SELECT GETDATE() StartDate, DATEADD(DAY, 11, GETDATE()) EndDate) a

这对表中的多个开始日期和结束日期很有用。不需要变量和循环！

工作原理：确定开始日期和结束日期之间的整周数

确定不到一周的天数，并添加开始日期索引（1,2,3,4,5,6,7），如果该值大于计数8（表示星期日！）

对于周日周日指数为1，第8天只是指数1。

Answer 5

试试这个

select count(DAYNAME(your_date_field)='Sunday') as sunday from tbl_name where your_date_field between 'date1' AND 'date2'

Answer 6

SELECT COUNT(*) FROM table_name
WHERE column BETWEEN 'start_date' AND 'end_date'
AND WEEKDAY(date) = 6;

Answer 7

参考这个，mysql有一个函数dayofweek，你可以很容易地计算开始的工作日和之后用7分的天数：

http://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html#function_dayofweek

OR

SELECT COUNT(*) FROM table_name
WHERE date_column_of_table_name BETWEEN '2013-01-11' AND '2013-03-01'
AND WEEKDAY(date_column_of_table_name) = 6

如何使用mysql计算两个日期之间的星期日数

7 个答案: