Question

我尝试过搜索，但我可能会使用错误的关键字，因为我找不到答案。

我正在尝试查找两个日期和员工之间打开的订单数量。我有一个表显示员工列表，另一个表显示包含打开和关闭日期的订单列表，如果有帮助，还显示日期表。

加入的员工和订单表将返回如下内容：

employee    order ref   opened          closed
a           123         01/01/2012      04/01/2012
b           124         02/01/2012      03/01/2012
a           125         02/01/2012      03/01/2012

我需要将这些数据转换为：

Date            employee    Count
01/01/2012      a           1
02/01/2012      a           2
02/01/2012      b           1
03/01/2012      a           2
03/01/2012      b           1
04/01/2012      a           1

我从SQL服务器中提取数据。

有什么想法吗？

由于

尼克

Answer 1

将Dates加入Employees和Orders之间的联接结果，然后按日期和员工分组以获取计数，如下所示：

SELECT
  d.Date,
  o.Employee,
  COUNT(*) AS count
FROM Employees e
  INNER JOIN Orders o ON e.ID = o.Employee
  INNER JOIN Dates d ON d.Date BETWEEN o.Opened AND o.Closed
GROUP BY
  d.Date,
  o.Employee

Answer 2

SELECT opened,employee,count(*)
FROM employee LEFT JOIN orders
WHERE opened < firstDate and opened > secondDate
GROUP BY opened,employee

或者你可以改变

中的第一个条件

WHERE opened BETWEEN firstDate and secondDate

Answer 3

调用结果列计数有点奇怪，因为它实际上似乎是一个行号。您可以使用ROW_NUMBER执行此操作。

另一个有趣的部分是您还希望打开日期和关闭日期作为单独的行。使用简单的UNION将解决这个问题。

WITH cte 
     AS (SELECT Row_number() OVER ( PARTITION BY employee  
                                    ORDER BY order_ref) count, 
                employee, 
                opened, 
                closed 
         FROM   orders) 
SELECT employee,  opened date,  count 
FROM   cte 
UNION ALL 
SELECT employee,  closed date,  count 
FROM   cte 
ORDER  BY Date, 
          employee

DEMO

Answer 4

我最喜欢的方法是计算累计开盘次数和累计收盘次数。

with cumopens as
    (select employee, opened as thedate,
            row_number() over (partition by employee order by opened) as cumopens,
            0 as cumcloses
     from eo
    ),
     cumcloses as
    (select employee, closed as thedate, 0 as cumopens,
            row_number() over (partition by employee order by closed ) as cumcloses
     from eo
    )
select employee, c.thedate, max(cumopens), max(cumcloses),
       max(cumopens) - max(cumcloses) as stillopened
from ((select *
       from cumopens
      ) union all
      (select *
       from cumcloses
      )
     ) c
group by employee, thedate

这种方法的唯一问题是只报告有员工活动的日期。这适用于您的情况。

更通用的解决方案需要序列号来生成日期。为此，我经常从一些具有足够行的现有表中创建一个：

with nums as
    (select row_number() over (partition by null order by null) as seqnum
     from employees
    )
select employee, dateadd(day, opened, seqnum) as thedate, count(*)
from eo join
     nums
     on datediff(day, opened, closed) < seqnum
group by employee, dateadd(day, opened, seqnum)
order by 1, 2

两个日期之间每天开放订单的SQL计数

4 个答案: