20

我需要计算两个日期之间的天数（工作日），不包括周末（最重要）和节假日

SELECT DATEDIFF(end_date, start_date) from accounts

但是，我不知道我应该如何在 MySQL 中做到这一点，我发现这篇文章Count days between two dates，不包括周末（仅限 MySQL）。我无法弄清楚如何在 mysql 中进行功能查询，您能否提供一些有关如何使用 mysql 查询来实现这一点的信息。如果我缺少任何东西，请告诉我。

[编辑]

CREATE TABLE `candidatecase` (
  `ID` int(11) NOT NULL AUTO_INCREMENT COMMENT 'Unique ID',
  `CreatedBy` int(11) NOT NULL,
  `UseraccountID` int(11) NOT NULL COMMENT 'User Account ID',
  `ReportReadyID` int(11) DEFAULT NULL COMMENT 'Report Ready ID',
  `DateCreated` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP COMMENT 'Date Created',
  `InitiatedDate` timestamp NULL DEFAULT '0000-00-00 00:00:00' COMMENT 'Date Initiated',
  `ActualCompletedDate` timestamp NULL DEFAULT '0000-00-00 00:00:00' COMMENT 'Date Completed Case',
  `ProjectedCompletedDate` timestamp NULL DEFAULT '0000-00-00 00:00:00' COMMENT 'Date Projected Finish',
  `CheckpackagesID` int(11) DEFAULT NULL COMMENT 'Default Check Package Auto Assign Once Initiate Start',
  `Alacartepackage1` int(11) DEFAULT NULL COMMENT 'Ala carte Request #2',
  `Alacartepackage2` int(11) DEFAULT NULL COMMENT 'Ala carte Request #3',
  `OperatorID` int(11) NOT NULL COMMENT 'User Account - Operator',
  `Status` int(11) NOT NULL COMMENT 'Status',
  `caseRef` varchar(100) NOT NULL,
  PRIMARY KEY (`ID`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1 AUTO_INCREMENT=293 ;

--
-- Dumping data for table `candidatecase`
--

INSERT INTO `candidatecase` (`ID`, `CreatedBy`, `UseraccountID`, `ReportReadyID`, `DateCreated`, `InitiatedDate`, `ActualCompletedDate`, `ProjectedCompletedDate`, `CheckpackagesID`, `Alacartepackage1`, `Alacartepackage2`, `OperatorID`, `Status`, `caseRef`) VALUES
(1, 43, 70, NULL, '2011-07-22 02:29:31', '2011-07-07 07:27:44', '2011-07-22 02:29:31', '2011-07-17 06:53:52', 11, NULL, NULL, 44, 6, ''),
(2, 43, 74, NULL, '2012-04-03 04:17:15', '2011-07-11 07:07:23', '2011-07-13 05:32:58', '2011-07-21 07:01:34', 20, 0, 0, 51, 0, ''),
(3, 43, 75, NULL, '2011-07-29 04:10:07', '2011-07-11 07:27:12', '2011-07-29 04:10:07', '2011-07-21 07:02:14', 20, NULL, NULL, 45, 6, ''),
(4, 43, 78, NULL, '2011-07-18 03:32:27', '2011-07-11 07:51:31', '2011-07-13 02:18:34', '2011-07-21 07:37:53', 20, NULL, NULL, 45, 6, ''),
(5, 43, 76, NULL, '2011-07-29 04:09:19', '2011-07-11 07:51:11', '2011-07-29 04:09:19', '2011-07-21 07:38:30', 20, NULL, NULL, 45, 6, ''),
(6, 43, 77, NULL, '2011-07-18 03:32:49', '2011-07-11 07:51:34', '2011-07-18 02:18:46', '2011-07-21 07:39:00', 20, NULL, NULL, 45, 6, ''),
(7, 43, 79, NULL, '2011-07-18 03:33:02', '2011-07-11 07:53:24', '2011-07-18 01:50:12', '2011-07-21 07:42:57', 20, NULL, NULL, 45, 6, ''),
(8, 43, 80, NULL, '2011-07-29 04:10:38', '2011-07-11 07:53:58', '2011-07-29 04:10:38', '2011-07-21 07:43:14', 20, NULL, NULL, 45, 6, ''),
(9, 43, 81, NULL, '2011-07-18 03:31:54', '2011-07-11 07:53:49', '2011-07-13 02:17:02', '2011-07-21 07:43:43', 20, NULL, NULL, 45, 6, ''),
(11, 43, 88, NULL, '2011-07-18 03:15:53', '2011-07-13 04:57:38', '2011-07-15 08:57:15', '2011-07-23 04:39:14', 12, NULL, NULL, 44, 6, ''),
(13, 43, 90, NULL, '2011-07-26 07:39:24', '2011-07-13 12:16:48', '2011-07-26 07:39:24', '2011-07-23 12:13:50', 15, NULL, NULL, 51, 6, ''),
(63, 43, 176, NULL, '2011-09-13 08:23:13', '2011-08-26 10:00:32', '2011-09-13 08:23:13', '2011-09-05 09:58:47', 41, NULL, NULL, 45, 6, ''),
(62, 43, 174, NULL, '2011-08-24 03:54:30', '2011-08-24 03:53:13', '2011-08-24 03:54:30', '2011-08-29 03:52:48', 17, NULL, NULL, 51, 6, ''),
(61, 43, 173, NULL, '2011-08-24 03:55:05', '2011-08-24 03:53:39', '2011-08-24 03:55:05', '2011-08-29 03:52:36', 17, NULL, NULL, 51, 6, ''),
(60, 43, 172, NULL, '2011-08-24 03:22:41', '2011-08-24 03:21:50', '2011-08-24 03:22:41', '2011-08-29 03:21:11', 17, NULL, NULL, 51, 6, ''),
(59, 43, 171, NULL, '2011-08-24 03:23:19', '2011-08-24 03:22:00', '2011-08-24 03:23:19', '2011-08-29 03:20:57', 17, NULL, NULL, 51, 6, '');
4

5 回答 5

24

你可能想试试这个：

  1. 计算工作日数（从这里获取）

    SELECT 5 * (DATEDIFF('2012-12-31', '2012-01-01') DIV 7) + MID('0123444401233334012222340111123400012345001234550', 7 * WEEKDAY('2012-01-01') + WEEKDAY('2012-12-31') + 1, 1)

    这为您提供了 2012 年的 261 个工作日。

  2. 现在您需要知道不是周末的假期

    SELECT COUNT(*) FROM holidays WHERE DAYOFWEEK(holiday) < 6

    其结果取决于您的假期表。

  3. 我们需要在一个查询中得到它：

    SELECT 5 * (DATEDIFF('2012-12-31', '2012-01-01') DIV 7) + MID('0123444401233334012222340111123400012345001234550', 7 * WEEKDAY('2012-01-01') + WEEKDAY('2012-12-31') + 1, 1) - (SELECT COUNT(*) FROM holidays WHERE DAYOFWEEK(holiday) < 6)

    这应该是它。

编辑：请注意，只有当您的结束日期高于开始日期时，这才能正常工作。

于 2012-04-26T10:09:41.570 回答
4

创建一个表，其中包含未来 100 年的所有周末和节假日。

鉴于没有人知道 2052 年的假期是什么，您需要能够指定一天是什么时候是“假期”，无论如何此时您将无法做出准确的功能。只需在每年知道假期时更新您的非工作日表（但您总是会知道周末）。

然后您的查询变为：

SELECT DATEFIFF(end_date, start_date) - COALESCE((SELECT COUNT(1) FROM nonWorkDays WHERE nonWorkDays.date BETWEEN start_date AND end_date), 0)
FROM accounts

如果你真的需要编写一个DATEDIFFWITHOUTWEEKENDSORHOLIDAYS函数，那么只需使用上面的方法并创建一个函数（关于如何在每个 RDBMS 中创建函数的资源很多）.. 一定要给它一个更好的名字。^_^

您需要解决的一件事是我认为上面某处缺少 +1，例如 DATEDIFF(today, today) 如果今天是周末将返回 -1 而不是返回 0。

于 2012-04-26T10:06:01.867 回答
1

这样的事情可能会奏效。将所有假期日期和周末日期添加到表中。

SELECT 
  DATEDIFF(end_date, start_date) 
FROM table
WHERE date NOT IN (SELECT date FROM holidaydatestable )
于 2012-04-26T10:15:33.740 回答
1

试试这个代码，这将计算不包括周末的天数

 SELECT
       (DATEDIFF(dd, @StartDate, @EndDate)+1)
      -(DATEDIFF(wk, @StartDate, @EndDate) * 2)
from test_tbl where date NOT IN (SELECT date FROM holidaydatestable )
于 2014-08-14T10:49:21.933 回答
0

制作一个函数，该函数将在日期之间进行一段时间循环，以增加不是星期六或星期日的天数。

于 2012-04-26T09:47:31.040 回答