两个日期之间每天未结订单的 SQL 计数答案

【问题标题】：SQL Count Of Open Orders Each Day Between Two Dates两个日期之间每天未结订单的 SQL 计数
【发布时间】：2012-06-26 14:23:19
【问题描述】：

我尝试过搜索，但我可能使用了错误的关键字，因为我找不到答案。

我正在尝试查找在两个日期之间以及按员工打开的订单数量。我有一个显示员工列表的表，另一个显示包含打开和关闭日期的订单列表以及日期表（如果有帮助的话）。

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成：

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

我正在从 SQL 服务器中提取数据。

有什么想法吗？

谢谢

尼克

【问题讨论】：

如果根本没有行，你需要它显示零吗？

标签： sql sql-server

【解决方案1】：

将Dates 加入Employees 和Orders 之间的连接结果，然后按日期和员工分组以获得计数，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

【讨论】：

【解决方案2】：

我最喜欢的方法是计算累积打开次数和累积关闭次数。

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一个序列号来生成日期。为此，我经常从一些具有足够行数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

【讨论】：

不小心被否决了，现在我无法撤消了。对不起。

【解决方案3】：

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

或者你可以改变第一个条件

WHERE opened BETWEEN firstDate and secondDate

【讨论】：

【解决方案4】：

调用结果列计数有点奇怪，因为它实际上似乎是一个行号。您可以通过使用 ROW_NUMBER 来做到这一点。

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。使用一个简单的 UNION 就可以解决这个问题。

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

【讨论】：