T-Sql每年獲取員工的開始和結束日期

Question

我有一個名為“ Employee ”的表

員工表具有以下列

Id (Identity)
EmploymentStartDate (datetime),
EmploymentEndDate (nullable datetime)

T-SQL查詢

DECLARE @FromYear int, @ToYear int

SELECT @FromYear = YEAR(MIN(EmploymentStartDate)),
       @ToYear = YEAR(GETDATE())
FROM Employee

;WITH CTE AS 
(
    SELECT @FromYear As TheYear
    UNION ALL
    SELECT TheYear + 1
    FROM CTE
    WHERE TheYear < @ToYear
)

SELECT TheYear as [Year], 
       COUNT
       (
       CASE WHEN TheYear <= YEAR(COALESCE(EmploymentEndDate, GETDATE())) THEN 
           1 
       END
       ) As [EmployeeCountPerYear]
FROM CTE
INNER JOIN Employee ON(TheYear >= YEAR(EmploymentStartDate))
GROUP BY TheYear

題：

Employee表具有下面的行數據。

Id - EmploymentStartDate  -  EmploymentEndDate
1  - '2012-10-10'         -   null
2  - '2014-10-10'         -   '2015-10-10'
3  - '2015-10-10'         -   null
4  - '2016-10-10'         -   null
5  - '2017-10-10'         -   null

根據上表的行值結果應如下所示在我的查詢中

TheYear - EmployeeCountPerYear

2012    -  1
2013    -  1
2014    -  2
2015    -  3 (Because EmploymentEndDate has one employee that worked in 2015)
2016    -  3
2017    -  4

但是，如果我運行查詢，我將看不到上述結果。我不確定我能否說出問題，但我想逐年查找所有員工的工作開始日期和結束日期 。我們將不勝感激。謝謝。

Answer 1

您可以嘗試OUTER APPLY ：

DECLARE @FromYear int, @ToYear int;

SELECT @FromYear = YEAR(MIN(EmploymentStartDate)),
       @ToYear = YEAR(GETDATE())
FROM Employee;

WITH CTE AS 
(
    SELECT @FromYear As TheYear
    UNION ALL
    SELECT TheYear + 1
    FROM CTE
    WHERE TheYear < @ToYear
)
SELECT *
FROM CTE A
OUTER APPLY(SELECT COUNT(*) EmploymentStartDate  
            FROM dbo.Employee
            WHERE A.TheYear 
            BETWEEN YEAR(EmploymentStartDate) AND YEAR(ISNULL(EmploymentEndDate,GETDATE()))) B;

這是一個演示 。 結果是：

╔═════════╦═════════════════════╗
║ TheYear ║ EmploymentStartDate ║
╠═════════╬═════════════════════╣
║    2012 ║                   1 ║
║    2013 ║                   1 ║
║    2014 ║                   2 ║
║    2015 ║                   3 ║
║    2016 ║                   3 ║
║    2017 ║                   4 ║
╚═════════╩═════════════════════╝