如何计算公司在不雇用任何人的情况下最长的工作天数

Question

鉴于下表，我想计算公司在不雇用任何人的情况下最长的工作天数。

<< img src =“ https://image.soinside.com/eyJ1cmwiOiAiaHR0cHM6Ly9pLnN0YWNrLmltZ3VyLmNvbS85QTVlSC5wbmcifQ==” alt =“表的图像”>

这是我到目前为止尝试过的：

SELECT EmpID, MAX(Hire_Date-e.Hire_Date) AS maxDiff 
FROM employees
    LEFT JOIN employees e ON e.EmpID = employees.EmpID 
        AND employees.Hire_Date >= e.Hire_Date
WHERE e.Hire_Date IS NOT NULL
GROUP BY EmpID;

Answer 1

您要使用lag()或lead()：

select datediff(next_hd, hiredate) as days_since_prev_hire
from (select t.*,
             lead(hiredate) over (order by hiredate) as next_hd
      from t
     ) t
where next_hd is not null
order by datediff(next_hd, hiredate) desc;

请注意，这不包括最后的最新雇用。为此，删除where子句并使用：

order by datediff(coalesce(next_hd, curdate()), hiredate) desc

Answer 2

使用ANSI SQL并将字符串转换为日期：

SELECT
  empid,
  STR_TO_DATE(hire_date,'%d/%m/%Y') AS hire_date,
  DATEDIFF(
    STR_TO_DATE(hire_date,'%d/%m/%Y'),
    (SELECT MAX(STR_TO_DATE(hire_date,'%d/%m/%Y')) FROM hiring WHERE STR_TO_DATE(hire_date,'%d/%m/%Y') < STR_TO_DATE(t.hire_date,'%d/%m/%Y'))
  ) AS days_since_last_hiring
FROM hiring AS t ORDER BY STR_TO_DATE(hire_date,'%d/%m/%Y');

输出：

+-------+------------+------------------------+
| empID | hire_date  | days_since_last_hiring |
+-------+------------+------------------------+
|   103 | 2015-02-03 |                   NULL |
|   407 | 2016-01-07 |                    338 |
|   333 | 2016-08-07 |                    213 |
|   409 | 2016-10-15 |                     69 |
|   410 | 2017-01-01 |                     78 |
|   413 | 2017-01-01 |                     78 |
+-------+------------+------------------------+
6 rows in set (0.00 sec)

当然，最后两行与前一行具有较低日期的比较相比具有相同的差异