Question

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）和假期

SELECT DATEDIFF(end_date, start_date) from accounts

但是，我不知道我应该如何在MySQL中这样做，我发现了这篇文章Count days between two dates, excluding weekends (MySQL only)。我无法弄清楚如何在mysql中进行函数查询，能否给出一些如何用mysql查询实现这一点的信息。如果我遗失任何东西，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, '');

Answer 1

您可能想尝试一下：

计算工作天数（从here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。
现在你需要知道周末没有的假期

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

结果取决于您的假期表。
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。

修改：请注意，只有当您的结束日期高于开始日期时才能正常使用。

Answer 2

创建一个表格，其中包含下一个100年的所有周末和假日。

如果没有人知道2052年的假期，您需要能够指定某一天是“假日”，但无论如何，您将无法在此时制作准确的功能。每年假期结束时更新您的非工作日表（但您将始终知道周末）。

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函数并创建一个函数（关于如何在每个RDBMS中创建函数有很多资源）。只需确保给它一个更好的名字。 ^ _ ^

你需要解决的一件事是我认为上面某处有+1丢失，例如DATEDIFF（今天，今天），如果今天周末将返回-1而不是返回0。

Answer 3

这样的事可能有用。将所有假日日期和周末日期添加到表格中。

SELECT 
  DATEDIFF(end_date, start_date) 
FROM table
WHERE date NOT IN (SELECT date FROM holidaydatestable )

Answer 4

尝试此代码，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 )

Answer 5

创建一个函数，在不是星期六或星期日的日期之间进行一段时间的循环。

如何计算MySQL中不包括周末和假日的日期差异

5 个答案: