Sql 用于检索员工计数的单个查询

我有一个包含以下字段的employee表

员工（id、姓名、加入日期、工资、部门）

我想检索每个部门过去1天、2天、3天和4天内加入公司的所有员工人数。请在下面的链接中查看我需要的结果

样本结果：

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如果你想在过去的24小时内，在24到48小时之间，等等

SELECT dept,
       COUNT( CASE WHEN joiningDate BETWEEN SYSDATE - 1 AND SYSDATE     THEN 1 ELSE NULL END ) AS day1,
       COUNT( CASE WHEN joiningDate BETWEEN SYSDATE - 2 AND SYSDATE - 1 THEN 1 ELSE NULL END ) AS day2,
       COUNT( CASE WHEN joiningDate BETWEEN SYSDATE - 3 AND SYSDATE - 2 THEN 1 ELSE NULL END ) AS day3,
       COUNT( CASE WHEN joiningDate BETWEEN SYSDATE - 4 AND SYSDATE - 3 THEN 1 ELSE NULL END ) AS day4,
       COUNT( CASE WHEN joiningDate BETWEEN SYSDATE - 5 AND SYSDATE - 4 THEN 1 ELSE NULL END ) AS day5
FROM   Employee
GROUP BY dept;

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如果您昨天、2天前、3天前等需要它，请将

TRUNC（）

包装在

SYSDATE

的每个实例周围，以下内容将为您提供所需内容：

SELECT DEPT,
       SUM(CASE WHEN TRUNC(SYSDATE) - TRUNC(JOININGDATE) = 1 THEN 1 ELSE 0 END) AS day1,
       SUM(CASE WHEN TRUNC(SYSDATE) - TRUNC(JOININGDATE) = 2 THEN 1 ELSE 0 END) AS day2,
       SUM(CASE WHEN TRUNC(SYSDATE) - TRUNC(JOININGDATE) = 3 THEN 1 ELSE 0 END) AS day3,
       SUM(CASE WHEN TRUNC(SYSDATE) - TRUNC(JOININGDATE) = 4 THEN 1 ELSE 0 END) AS day4,
       SUM(CASE WHEN TRUNC(SYSDATE) - TRUNC(JOININGDATE) = 5 THEN 1 ELSE 0 END) AS day5
FROM EMPLOYEE
GROUP BY DEPT
ORDER BY DEPT;

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如果我们有pivot标签，为什么不使用

pivot

select * from (
    select dept, trunc(sysdate)-trunc(joiningDate) dt
      from employee where joiningDate >= trunc(sysdate)-5)
  pivot (count(1) for dt in (1 day1, 2 day2, 3 day3, 4 day4, 5 day5)) 
  order by dept

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请向我们展示您迄今为止所做的尝试。是否可以从计数中排除周末？类似于：

COUNT（在TRUNC（SYSDATE）之间连接日期时的大小写）-何时添加字符（SYSDATE，'D'）>1，然后1其他3结束，以及TRUNC（SYSDATE）-何时添加字符（SYSDATE，'D'）>1，然后0其他2结束，然后1其他空结束）但是，作为第1天，请注意
NLS_TERRITORY
将影响
'D'
参数返回的值-请参见此处：是否可以从计数中排除周末？我希望您可以这样做。