选择已打开，已关闭和待办事项

时间：2014-04-11 12:27:13

标签： sql sql-server select group-by

我有下表：

+-------------+---------------------+---------------------+
| status      | open_time           | close_time          |
+-------------+---------------------+---------------------+
|      closed | 01-11-2014 19:32:44 | 01-11-2014 20:32:44 |
|        open | 01-12-2014 22:33:49 | 02-12-2014 22:33:49 |
|        open | 01-23-2014 22:08:24 | 03-23-2014 22:08:24 |
|      closed | 02-01-2014 22:33:57 | 03-01-2014 22:33:57 |
|        open | 02-01-2013 22:37:34 | 02-01-2013 23:37:34 |
|      closed | 04-20-2013 15:23:00 | 05-20-2013 15:23:00 |
|        open | 04-20-2013 12:21:49 | 05-20-2013 12:21:49 |
|      closed | 04-25-2013 11:22:00 | 06-25-2013 11:22:00 |
|      closed | 05-20-2013 14:23:49 | 10-20-2013 14:23:49 |
|      closed | 04-20-2013 16:33:49 | 04-25-2013 16:33:49 |
+-------------+---------------------+---------------------*

我想展示这个结果：

+-------------+---------------+--------------+---------+
| Year | Month | Opened Cases | Closed Cases | Backlog |
+-------------+---------------+--------------+---------+
| 2014 |     4 |           10 |            5 |      62 | (57 + 5)
| 2014 |     3 |            9 |            7 |      57 | (52 + 2)    
| 2014 |     2 |           15 |           20 |      52 | (57 - 5)
| 2014 |     1 |           12 |            1 |      57 | (46 + 11)
| 2013 |    12 |           10 |            9 |      46 | (45 + 1)
| 2013 |    11 |           50 |            5 |      45 | (45)
+--------------+--------------+--------------+---------+

其实我面临两种情况：

  • 情况＃1：我无法为已结案例获得正确的价值。试图在没有运气的地方使用条款。

  • 情况＃2：在关闭已关闭的案例后，我应该做（已打开 - 已关闭）并在几个月内累积，因此这将是积压案件。

对于情境＃1：

如果我执行以下操作，请选择：

SELECT
  YEAR(open_time) AS Ano,
  MONTH(open_time) AS Mes,
  sum(CASE WHEN DATEPART(YYYY, open_time)= 2013 AND DATEPART(MM, open_time)= 11 THEN 1 ELSE 0 END) Abertos,
  sum(CASE WHEN DATEPART(YYYY, close_time)= 2013 AND DATEPART(MM, close_time)= 11 THEN 1 ELSE 0 END) Fechados
FROM
  TABLE
WHERE
  GROUPDESC= 'SUPPORT'
GROUP BY
  MONTH(open_time),
  YEAR(open_time)
ORDER BY
  Ano DESC,
  Mes DESC

我得到（这是关闭案例的正确值）：

+-------------+---------------+--------------+
| Year | Month | Opened Cases | Closed Cases |
+-------------+---------------+--------------+
| 2014 |     4 |            0 |            0 |
| 2014 |     3 |            0 |            0 |
| 2014 |     2 |            0 |            0 |
| 2014 |     1 |            0 |            0 |
| 2013 |    12 |            0 |            0 |
| 2013 |    11 |           50 |            5 |
+--------------+--------------+--------------+

但如果我这样做：

SELECT
  YEAR(open_time) AS Ano,
  MONTH(open_time) AS Mes,
  sum(CASE WHEN DATEPART(YYYY, open_time)= YEAR(open_time) AND DATEPART(MM, open_time)= MONTH(open_time) THEN 1 ELSE 0 END) Abertos,
  sum(CASE WHEN DATEPART(YYYY, close_time)= YEAR(close_time) AND DATEPART(MM, close_time)= YEAR(close_time) THEN 1 ELSE 0 END) Fechados
FROM
  TABLE
WHERE
  GROUPDESC= 'SUPPORT'
GROUP BY
  MONTH(open_time),
  YEAR(open_time)
ORDER BY
  Ano DESC,
  Mes DESC

我明白了：

+-------------+---------------+--------------+
| Year | Month | Opened Cases | Closed Cases |
+-------------+---------------+--------------+
| 2014 |     4 |            0 |            0 |
| 2014 |     3 |            0 |            0 |
| 2014 |     2 |            0 |            0 |
| 2014 |     1 |            0 |            0 |
| 2013 |    12 |            0 |            0 |
| 2013 |    11 |           50 |           50 |
+--------------+--------------+--------------+

2 个答案:

答案 0 :(得分：1)

以下是解决问题的一种方法：选择每月的所有开仓数和每月的所有结算数。然后全外连接它们。累积是一个所谓的运行总计，你可以使用SUM OVER。

select
  coalesce(opened.ano, closed.ano) as ano,
  coalesce(opened.mes, closed.mes) as mes,
  coalesce(opened.cnt, 0) as opened_cases,
  coalesce(closed.cnt, 0) as closed_cases,
  sum(coalesce(opened.cnt, 0) - coalesce(closed.cnt, 0)) over (order by coalesce(opened.ano, closed.ano), coalesce(opened.mes, closed.mes)) as backlog
from
(
  select 
    year(open_time) as ano, 
    month(open_time) as mes,
    count(*) as cnt
  from probsummarym1
  where groupdesc = 'SUPPORT'
  group by year(open_time), month(open_time)
) opened
full outer join
(
  select 
    year(close_time) as ano, 
    month(close_time) as mes,
    count(*) as cnt
  from probsummarym1
  where groupdesc = 'SUPPORT'
  and status = 'closed'
  group by year(close_time), month(close_time)
) closed
  on opened.ano = closed.ano and opened.mes = closed.mes
order by coalesce(opened.ano, closed.ano) desc, coalesce(opened.mes, closed.mes) desc;

这是SQL小提琴：http://sqlfiddle.com/#!6/68dcf/7。

另一种方法是使用UNION ALL粘合开始事件和关闭事件，然后计数：

select 
  ano,
  mes,
  opened_cases,
  closed_cases,
  sum(opened_cases - closed_cases) over (order by ano, mes) as backlog
from
(
  select 
    year(fecha) as ano,
    month(fecha) as mes,
    sum(case when evento = 'opened' then 1 else 0 end) as opened_cases,
    sum(case when evento = 'closed' then 1 else 0 end) as closed_cases
  from
  (
    select 'opened' as evento, open_time as fecha
    from probsummarym1
    where groupdesc = 'SUPPORT'
    union all
    select 'closed' as evento, close_time as fecha
    from probsummarym1
    where groupdesc = 'SUPPORT'
    and status = 'closed'
  ) x 
  group by year(fecha), month(fecha)
) y
order by ano desc, mes desc;

这是第二个SQL小提琴：http://sqlfiddle.com/#!6/68dcf/18。

编辑：没有SUM？这太糟糕了。所以你必须再次选择计数。这很慢，因为必须每个月再次扫描一次表。

对于每个月，我们必须找到所有开始日期，直到那时为止所有结束日期。由于在开始之前永远不会发生结束，我们可以选择开始日期匹配的所有记录并计算这些记录。在这些记录中，我们还将找到所有可能的结束日期。我们统计这些（用和和案例），减去，我们就完成了。

所以你必须更换SUM() OVER as backlog部分：

  (
    select
      count(*)
      -
      sum
      (
        case 
          when eventsuntil.status = 'closed' 
          and year(eventsuntil.close_time) * 100 + month(eventsuntil.close_time) >= 
              y.ano * 100 + y.mes 
        then 1 else 0
        end
      )    
    from probsummarym1 eventsuntil
    where eventsuntil.groupdesc = 'SUPPORT'
    and year(eventsuntil.open_time) * 100 + month(eventsuntil.open_time) >= 
        y.ano * 100 + y.mes
  ) as backlog

但是，即使在2012版本中，SQL Server也无法执行此操作，我认为这是一个dbms漏洞。 y.ano和y.mes（或coalesce（opened.ano，closed.ano）和coalesce（opens.mes，closed.mes）的第一个语句）应被视为内部查询的常量，因为评估是按照每个外部记录，即月份，但它们不是。我不知道如何克服这个问题。也许SQL Server的一些专家可以帮助你。

以下是导致语法错误的两个小提琴：http://sqlfiddle.com/#!6/68dcf/32和http://sqlfiddle.com/#!6/68dcf/31。对不起，我不能再帮到你了。

答案 1 :(得分：0)

试试这个。

;with CTE as (
    SELECT
        row_number() over (partition by groupdesc order by YEAR(open_time), MONTH(open_time)) "Date_Order",
        YEAR(open_time) "Year",
        MONTH(open_time) "Month",
        SUM(CASE status WHEN 'open' THEN 1 ELSE 0 END) "open_count",
        SUM(CASE status WHEN 'closed' THEN 1 ELSE 0 END) "closed_count",
        SUM(CASE status WHEN 'open' THEN 1 ELSE 0 END) - SUM(CASE status WHEN 'closed' THEN 1 ELSE 0 END) "monthly_change"
    FROM PROBSUMMARYM1
    WHERE groupdesc = 'SUPPORT'
    GROUP BY groupdesc,
        YEAR(open_time),
        MONTH(open_time)
)
select c1.Date_Order,
    c1.Year,
    c1.Month,
    c1.open_count,
    c1.closed_count,
    sum(c2.monthly_change) "Backlog"
from CTE c1
join CTE c2
    on c2.Date_Order <= c1.Date_Order
group by c1.Date_Order,
    c1.Year,
    c1.Month,
    c1.open_count,
    c1.closed_count
order by c1.Year desc,
    c1.Month desc;

partition by groupdesc然后按groupdesc进行分组并非绝对必要，但它演示了如何使用row_number()同时运行一个查询来为多个执行相同的操作groupdesc。

您可以使用SUM() OVER()来逃避，但我对OVER()条款没有经验。请参阅http://technet.microsoft.com/en-us/library/ms189461.aspx，特别是示例C。

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