SQL選擇每天的最早開始日期時間和最晚結束日期時間

Question

我有一個下表，用於跟蹤員工的每日入職情況。 員工一天可以有多個簽入和簽出。 有沒有一種顯示方式，使得每個記錄都有EmployeeID以及每天的最早的CheckIn和最晚的CheckOut來匯總記錄。

CREATE TABLE IF NOT EXISTS `PhyMeS_schema`.`DailyEntry` (
  `EmployeeID` VARCHAR(10) NOT NULL,
  `CheckIn` DATETIME NOT NULL,
  `CheckOut` DATETIME NULL,
  PRIMARY KEY (`EmployeeID`, `CheckIn`),
  CONSTRAINT `fk_dailyEntry_EmployeeID`
    FOREIGN KEY (`EmployeeID`)
    REFERENCES `PhyMeS_schema`.`Employee` (`EmployeeID`)
    ON DELETE NO ACTION
    ON UPDATE NO ACTION)
ENGINE = InnoDB;

Answer 1

您應該能夠對EmployeeID和DATE(CheckIn)使用簡單的GROUP BY來做到這一點，如下所示：

SELECT
    EmployeeID
,   DATE(CheckIn) AS the_date
,   MIN(CheckIn) AS min_checkin
,   MAX(CheckOut) AS max_checkout
FROM DailyEntry
GROUP BY EmployeeID, DATE(CheckIn)

注意：此查詢假定所有簽出均與相應的簽入在同一天進行。 換句話說，午夜沒人在里面。 如果您需要處理夜班，查詢將變得更加復雜：

SELECT EmployeeID, the_date, MIN(min_checkin), MAX(max_checkout)
FROM (
    SELECT EmployeeID, DATE(CheckIn) AS the_date, MIN(CheckIn) AS min_checkin, null as max_checkout
    FROM DailyEntry
    GROUP BY EmployeeID, DATE(CheckIn)
UNION ALL
    SELECT EmployeeID, DATE(CheckOut) AS the_date, null AS min_checkin, MAX(CheckOut) as max_checkout
    FROM DailyEntry
    GROUP BY EmployeeID, DATE(CheckOut)
) GROUP BY EmployeeID, the_date