Question

我有一个表格数据，如何找到每个月的天数：

   serial      |      date_from      |      date_from
   001         |      2012-12-20     |       2013-01-25
   002         |      2012-12-20     |       2013-01-25
   003         |      2012-12-20     |       2013-01-25
   001         |      2013-01-26     |       2013-04-26

第一行的必需结果：

   Serial        |    days  |     month 
   001           |    11    |      December, 2012
   001           |    25    |      January, 2012

我怎样才能在mysql中执行此操作？

Answer 1

我有一个名为calendar的实用程序表，它存储了一个人可能遇到的所有日期（它是一张令人沮丧的小桌子）......

DROP TABLE IF EXISTS my_table;

CREATE TABLE my_table 
(serial      INT NOT NULL AUTO_INCREMENT PRIMARY KEY
,date_from DATE NOT NULL
,date_to DATE NULL
);

INSERT INTO my_table VALUES
(001,'2012-12-20','2013-01-25'),
(002,'2012-12-20','2013-01-25'),
(003,'2012-12-20','2013-01-25'),
(004,'2013-01-26','2013-04-26');


SELECT serial
     , DATE_FORMAT(dt,'%Y%m')
     , COUNT(*) total
  FROM calendar c 
  JOIN my_table x 
    ON c.dt > x.date_from 
   AND c.dt <= date_to 
 GROUP 
    BY serial
     , DATE_FORMAT(dt,'%Y%m');

+--------+------------------------+-------+
| serial | DATE_FORMAT(dt,'%Y%m') | total |
+--------+------------------------+-------+
|      1 | 201212                 |    11 |
|      1 | 201301                 |    25 |
|      2 | 201212                 |    11 |
|      2 | 201301                 |    25 |
|      3 | 201212                 |    11 |
|      3 | 201301                 |    25 |
|      4 | 201301                 |     5 |
|      4 | 201302                 |    28 |
|      4 | 201303                 |    31 |
|      4 | 201304                 |    26 |
+--------+------------------------+-------+

......或类似的东西

Answer 2

这有点棘手，你需要另一个行来源加入。我将使用内联视图作为示例，或者这可能来自一个表..

SELECT  t.service, expr, DATE_FORMAT(m._month,'%Y-%m')
  FROM ( SELECT '2012-12-01' AS _month 
          UNION ALL SELECT '2013-01-01'
          UNION ALL SELECT '2013-02-01'
          UNION ALL SELECT '2013-03-01'
          UNION ALL SELECT '2013-04-01'
       ) m
  JOIN mytable t
    ON ...

 WHERE ...

棘手的部分是正确获取JOIN的表达式（检查日期范围与月份的任何重叠），并使表达式返回天数。

这样的东西应该返回指定的结果：

SELECT d.serial
     , DATEDIFF(LEAST(d.date_to+INTERVAL 1 DAY,m._month+INTERVAL 1 MONTH)
               ,GREATEST(d.date_from,m._month)
       ) AS `days`
     , DATE_FORMAT(m._month,'%M, %Y') AS `month`
   --  , d.date_from
   --  , d.date_to
   --  , m._month
  FROM daterange d
  JOIN ( SELECT '2012-11-01' AS _month 
          UNION ALL SELECT '2012-12-01'
          UNION ALL SELECT '2013-01-01'
          UNION ALL SELECT '2013-02-01'
          UNION ALL SELECT '2013-03-01'
          UNION ALL SELECT '2013-04-01'
          UNION ALL SELECT '2013-05-01'
       ) m
    ON m._month >= DATE_FORMAT(d.date_from,'%Y-%m-01')
   AND m._month <= DATE_FORMAT(d.date_to  ,'%Y-%m-01')
 ORDER BY d.serial, m._month

Answer 3

我会计算下个月的第一天，然后计算两个日期之间的差异。

SELECT Serial, 
    DATEDIFF(DATE_FORMAT(date_from + INTERVAL 1 MONTH, '%Y-%m-01'), date_from)-1 AS `days`,
    DATE_FORMAT(date_from, '%M, %Y') AS `month`
FROM your_table_name

如何在mysql中查找日期范围内每个月的天数？

3 个答案: