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,我不知道该如何在MySQL中进行操作,我发现本文计算了两个日期之间的天数,不包括周末(仅MySQL)。我无法弄清楚如何在mysql中进行功能查询,能否提供一些有关如何使用mysql查询实现此功能的信息。如果我想念什么,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, '');
您可能要尝试以下操作:
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261个工作日(2012年)。
SELECT COUNT(*) FROM holidays WHERE DAYOFWEEK(holiday) < 6
其结果取决于您的假期表。
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)
应该是这样。
编辑: 请注意,这仅在结束日期高于开始日期时才能正常工作。