在 SQL 中生成每日总和的最有效方法？

如何解决在 SQL 中生成每日总和的最有效方法？

我正在开发一个包含多个组件的管理系统，其中一个是物理部件的管理。

在 SQL 中，我有一个零件转移表，说明他们要去哪里，他们来自哪里，以及转移发生的日期。与此相关的是每个转移的表格，其中包含转移的每个部分的条目以及金额。这些部分被连接到一个单独的表中，表明它们的类型。

我正在尝试构建一个报告，显示：自第一次零件转移以来的每一天，当天每个位置当前每种零件类型的数量是多少。我有一些 SQL 可以在我们较新的站点上很好地运行，但对于我们运行时间较长的站点（2017 年），由于零件/位置/天数如此之多，因此需要很长时间。

试图优化它以使其在大量数据上运行是愚蠢的差事吗？我正在寻找推进此工作的最佳方式，无论是开始构建某种记录缓存，还是在最有效的计算方式方面我遗漏了什么。

作为参考，以下是从我们最大的网站中提取的一些数据：

数据	计数
地点	20496
零件类型	224
天	1176

这里是大部分相关的 SQL，为了清楚起见，我确实清除了一些部分，但如果有遗漏，请告诉我，我会重新添加。

SELECT @earliestTransaction = min(actual_date) from partlist
WHERE (partlist.item_type = 'scaffold'
 AND partlist.partlist_type_id = 15)
 OR partlist.partlist_type_id = 6

SELECT @today = (select cast(getdate() as Date))

INSERT INTO @relevantScaffolds
SELECT s.id 
from scaffold s
join sitearea wa on wa.id = s.sitearea_id
where s.project_id = @ProjectID
AND (
(@siteAreaID IS NULL AND @scaffoldID IS NULL)
OR (@siteAreaID IS NOT NULL AND wa.parent_sitearea_id = @siteAreaID)
OR (@scaffoldID IS NOT NULL AND s.id = @scaffoldID)
)

WITH DateTable AS
(
    SELECT @earliestTransaction AS [DATE]
    UNION ALL
    SELECT DATEADD(dd,1,[DATE])
    FROM DateTable
    WHERE DATEADD(dd,[DATE]) <= @today
) 

--Get Date collection for all parts
INSERT INTO @datesAndParts
select DISTINCT
 dt.DATE,pc.id,rs.scaffold_id
from DateTable dt
join part_catalog pc on 1=1
join @relevantScaffolds rs on 1=1
--Remove limit for date recurison
OPTION (MaxRecursion 0)

WITH scaffold_transactions(part_catalog_id,quantity,actual_date,project_id,scaffold_id) AS
(
 -- Into Scaffold 
 SELECT part_catalog_id,sum(quantity),partlist.actual_date,partlist.project_id,rs.scaffold_id
 FROM partlist
  INNER JOIN partlist_part on partlist_part.partlist_id = partlist.id
  left join partlist parent on parent.id = partlist.parent_partlist_id
  join part_catalog on part_catalog.id = partlist_part.part_catalog_id
  join @relevantScaffolds rs on rs.scaffold_id = partlist.item_id
 WHERE partlist.item_type = 'scaffold'
 AND partlist.partlist_type_id = 15
 AND partlist.project_id = @ProjectID
 GROUP BY part_catalog_id,rs.scaffold_id
 
 UNION ALL
 -- deliveries_to_scaffold 
 SELECT part_catalog_id,SUM(quantity),rs.scaffold_id
 FROM partlist
  INNER JOIN workorder on workorder.id = partlist.item_id
  join @relevantScaffolds rs on rs.scaffold_id = workorder.scaffold_id
  INNER JOIN partlist_part on partlist_part.partlist_id = partlist.id
  left join partlist parent on parent.id = partlist.partsource_partlist_id
  join part_catalog on part_catalog.id = partlist_part.part_catalog_id

 WHERE partlist.partlist_type_id = 6
 AND partlist.project_id = @ProjectID
 GROUP BY part_catalog_id,rs.scaffold_id

 UNION ALL
 -- leaving_scaffold (negated quantities)
 SELECT part_catalog_id,-SUM(quantity),transferPartlist.actual_date,transferPartlist.project_id,rs.scaffold_id
 FROM partlist transferPartlist
  INNER JOIN partlist scaffoldPartlist on scaffoldPartlist.id = transferPartlist.parent_partlist_id
  INNER JOIN partlist_part on partlist_part.partlist_id = transferPartlist.id
  join part_catalog on part_catalog.id = partlist_part.part_catalog_id
  join @relevantScaffolds rs on rs.scaffold_id = scaffoldPartlist.item_id
 WHERE transferPartlist.partlist_type_id = 15
 AND scaffoldpartlist.partlist_type_id = 9
 AND scaffoldpartlist.project_id = @ProjectID
 GROUP BY part_catalog_id,rs.scaffold_id
)
INSERT INTO @keyedTransfers
select part_catalog_id,scaffold_id from scaffold_transactions
group by  part_catalog_id,scaffold_id

INSERT INTO @cleanupTable
select 
dap.date as date,dap.part_catalog_id,dap.scaffold_id,COALESCE(SUM(quantity) over (partition by dap.part_catalog_id,dap.scaffold_id order by dap.date ),0) as balance
from @datesAndParts dap
left join @keyedTransfers kt on kt.actual_date = dap.date and kt.part_catalog_id = dap.part_catalog_id and kt.scaffold_id = dap.scaffold_id

select date,pc.part_number,pc.description,scaffold.tag,balance
from @cleanupTable ct
join part_catalog pc on pc.id = ct.part_catalog_id
join scaffold on scaffold.id = ct.scaffold_id
where date >= @earliestTransaction AND date <= @today
and balance > 0
order by date,part_number

Here's a link to the execution plan 在成功的网站之一上运行时。我们在不同站点之间存在差距，它们都非常小或很大，中间没有，所以这个数据集要小得多。