如何在 MySQL 或 Google Data Studio 中按顺序条件分组？

Question

我有一个考勤数据库，员工记录存储如下

MySQL版本5.7.26-log（日期没有漏洞，所有日期都存在但为了方便，我已经从插入行语句中删除了批量日期数据）

CREATE TABLE `Whosebug` ( `id` int(9) NOT NULL DEFAULT '0', 
                               `Date` date NOT NULL, 
                               `EmpID` varchar(100) NOT NULL, 
                               `name` varchar(100) NOT NULL, 
                               `TeamName` varchar(100) NOT NULL, 
                               `Status` varchar(100) NOT NULL 
                             ) ENGINE=InnoDB DEFAULT CHARSET=latin1; 

-- -- Dumping data for table `Whosebug` -- 
INSERT INTO `Whosebug` (`id`, `Date`, `EmpID`, `name`, `TeamName`, `Status`) 
VALUES 
(5, '2019-03-01', '303016', 'Yatendra Ranawat', 'Computer Vision', 'P'), 
(23, '2019-03-01', '303128', 'Nikhat Khan', 'Media - 3D Automation', 'P'), 
(193606, '2019-09-02', '303016', 'Yatendra Ranawat', 'Computer Vision', 'P'), 
(194631, '2019-09-03', '303016', 'Yatendra Ranawat', 'Noon', '-'), 
(222309, '2019-09-30', '303016', 'Yatendra Ranawat', 'Noon', '-'), 
(223336, '2019-10-01', '303016', 'Yatendra Ranawat', 'Noon-Indore', 'P'), 
(282742, '2019-11-28', '303016', 'Yatendra Ranawat', 'Noon-Indore', '-'), 
(283765, '2019-11-29', '303016', 'Yatendra Ranawat', 'Home Depot - Indore', 'P'), 
(303251, '2019-12-18', '303128', 'Nikhat Khan', 'Media - 3D Automation', '-'), 
(304275, '2019-12-19', '303128', 'Nikhat Khan', 'US Taxonomy - Indore', 'P'), 
(309393, '2019-12-24', '303128', 'Nikhat Khan', 'US Taxonomy - Indore', 'P'), 
(310416, '2019-12-25', '303128', 'Nikhat Khan', 'Media - 3D Automation', 'PH'), 
(354076, '2020-02-06', '303016', 'Yatendra Ranawat', 'Home Depot - Indore', '-'), 
(354088, '2020-02-06', '303128', 'Nikhat Khan', 'Media - 3D Automation', 'P');

我正在使用 Google Data Studio 为团队中的员工持续时间创建报告，如下所示

当员工从一个团队调到另一个团队而不是 return 他以前工作过的团队时，它会完美地工作。但是，当一名员工 return 加入他之前工作过的团队时，它会聚集最大和最小逻辑

我需要在 MySQL / Google Data Studio 中做什么才能得到像下面这样的结果以及像上面那样的数据 table？

如果您需要我这边的任何详细信息，请告诉我。任何建议或指导将不胜感激。

Answer 1

测试

SELECT t4.EmpID, t4.name, t4.TeamName, MIN(t4.startdate) startdate, t4.enddate
FROM ( SELECT t1.EmpID, t1.name, t1.TeamName, t1.`Date` startdate, MAX(t2.`Date`) enddate
       FROM Whosebug t1
       JOIN Whosebug t2 ON t1.EmpID = t2.EmpID
                            AND t1.TeamName = t2.TeamName
                            AND t1.`Date` < t2.`Date`
       LEFT JOIN Whosebug t3 ON t1.EmpID = t3.EmpID
                                 AND t1.TeamName != t3.TeamName
                                 AND t1.`Date` < t3.`Date`                          
                                 AND t3.`Date` < t2.`Date`
       WHERE t3.EmpId IS NULL
       GROUP BY 1,2,3,4 ) t4
GROUP BY 1,2,3,5
ORDER BY 1,2,4,5;

= this query is too slow :( Server crashes while running on a table containing 222839 rows – Yatendra Ranawat

= @YatendraRanawat Convert LEFT JOIN into NOT EXISTS... – Akina

SELECT t4.EmpID, t4.name, t4.TeamName, MIN(t4.startdate) startdate, t4.enddate
FROM ( SELECT t1.EmpID, t1.name, t1.TeamName, t1.`Date` startdate, MAX(t2.`Date`) enddate
       FROM Whosebug t1
       JOIN Whosebug t2 ON t1.EmpID = t2.EmpID
                            AND t1.TeamName = t2.TeamName
                            AND t1.`Date` < t2.`Date`
       WHERE NOT EXISTS ( SELECT NULL
                          FROM Whosebug t3 
                          WHERE t1.EmpID = t3.EmpID
                            AND t1.TeamName != t3.TeamName
                            AND t1.`Date` < t3.`Date`                          
                            AND t3.`Date` < t2.`Date` )
       GROUP BY 1,2,3,4 ) t4
GROUP BY 1,2,3,5
ORDER BY 1,2,4,5;

fiddle

Answer 2

也测试下一个变体，速度更快：

SELECT EmpID, name, TeamName, startdate, MAX(enddate) enddate
FROM (
SELECT name,
       CASE WHEN EmpId = @id AND TeamName = @team 
            THEN @startdate
            ELSE @startdate := `Date` 
            END startdate,
       `Date` enddate,
       @id := EmpID EmpID,
       @team := TeamName TeamName
FROM Whosebug, (SELECT @id := '', @team := '', @startdate := '') variables
ORDER BY EmpID, `Date`
) t
GROUP BY 1,2,3,4
ORDER BY 1,2,4,5;

fiddle

在我的带有热缓存的系统上，它需要 0.03 秒，而上一个答案的查询分别为 25.39 秒和 1 分 54.79 秒。在您拥有 220k 条记录的系统上，差异一定更大。

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