为什么在此 SQL Server CTE 表达式中添加一行会更改结果答案

【问题标题】：Why does adding one line to this SQL Server CTE expression change result为什么在此 SQL Server CTE 表达式中添加一行会更改结果
【发布时间】：2015-04-10 12:24:37
【问题描述】：

我正在尝试创建 CTE（公用表表达式）来创建清单 BOM（材料清单）。 IE。查询遍历整个 BOM 并汇总每个唯一部分的数量。

我还需要返回一些关于 BOM 项的数据。

我正在尝试添加一个计算字段以显示 BOM 项是否为程序集，即它是否也有 bom 条目。我做错了什么？

数据库是 Aras Innovator，它使用 SQL Server。

记录结构是

<Part_Id> - <source_id : Bom Entry : related_id> - <Part_Id>

所以基本上有两个表。虽然查询还返回其他相关表中的数据。

第一个查询有效，但是当我尝试添加显示零件为装配的关系时，总和变为错误并返回错误的项目数。

<!-- language: lang-sql -->
WITH RecursiveBOM(source_id, related_id, quantity) AS
(
  SELECT pbom.SOURCE_ID, pbom.RELATED_ID, pbom.QUANTITY
    FROM innovator.PART_BOM AS pbom
    WHERE SOURCE_ID = '82FD83221C5946F0A70CADD328BDB6CA'

  UNION ALL

  SELECT pbom.SOURCE_ID, pbom.RELATED_ID, pbom.QUANTITY
    FROM innovator.PART_BOM AS pbom
    INNER JOIN RecursiveBOM rbom ON rbom.RELATED_ID = pbom.SOURCE_ID
)
SELECT
  rp.MAKE_BUY as MAKE_BUY,
  rp.ITEM_NUMBER AS PN,
  rp.NAME AS NAME,
  sum(RecursiveBOM.quantity) as Total,
  mp.ITEM_NUMBER as MPN,
  man.NAME as MANUFACTURER,
  vp.CATALOG_NUMBER as SPN,
  ven.NAME as SUPPLIER,      
  CAST(CASE WHEN count(cp.RELATED_ID)>0 THEN 'Yes' ELSE 'No' END AS TEXT) AS HAS_CAD,

FROM RecursiveBOM

INNER JOIN innovator.PART sp ON RecursiveBOM.source_id = sp.ID
INNER JOIN innovator.PART rp ON RecursiveBOM.related_id = rp.ID

--Manufacturer Part
LEFT JOIN innovator.PART_AML aml ON aml.SOURCE_ID = rp.ID
LEFT JOIN innovator.MANUFACTURER_PART mp ON aml.related_ID = mp.ID
LEFT JOIN innovator.MANUFACTURER man on mp.MANUFACTURER = man.ID

--Supplier
LEFT JOIN innovator.VENDOR_PART vp ON mp.ID = vp.RELATED_ID
LEFT JOIN innovator.VENDOR ven ON vp.SOURCE_ID = ven.ID

--Has drawings
LEFT JOIN innovator.CAD_PART cp ON cp.SOURCE_ID = rp.ID

--Don't decend into PCBA
WHERE sp.classification NOT LIKE '%PCBA%'

--Collect the results 
GROUP BY
  rp.ITEM_NUMBER,
  mp.ITEM_NUMBER,
  rp.NAME,
  man.NAME,
  vp.CATALOG_NUMBER,
  ven.NAME,
  cp.RELATED_ID,
  rp.MAKE_BUY

我只添加了两行，分别是

...
CAST(CASE WHEN count(bom.RELATED_ID)>0 THEN 'Yes' ELSE 'No' END as TEXT) AS IS_ASSY`
...
--Is Assembly
LEFT JOIN innovator.PART_BOM bom ON bom.SOURCE_ID = rp.ID
...

这会产生错误的总和：

<!-- language: lang-sql -->
WITH RecursiveBOM(source_id, related_id, quantity) AS
(
  SELECT pbom.SOURCE_ID, pbom.RELATED_ID, pbom.QUANTITY
    FROM innovator.PART_BOM AS pbom
    WHERE SOURCE_ID = '82FD83221C5946F0A70CADD328BDB6CA'

  UNION ALL

  SELECT pbom.SOURCE_ID, pbom.RELATED_ID, pbom.QUANTITY
    FROM innovator.PART_BOM AS pbom
    INNER JOIN RecursiveBOM rbom ON rbom.RELATED_ID = pbom.SOURCE_ID
)
SELECT
  rp.MAKE_BUY as MAKE_BUY,
  rp.ITEM_NUMBER AS PN,
  rp.NAME AS NAME,
  sum(RecursiveBOM.quantity) as Total,
  mp.ITEM_NUMBER as MPN,
  man.NAME as MANUFACTURER,
  vp.CATALOG_NUMBER as SPN,
  ven.NAME as SUPPLIER,      
  CAST(CASE WHEN count(cp.RELATED_ID)>0 THEN 'Yes' ELSE 'No' END AS TEXT) AS HAS_CAD,
  CAST(CASE WHEN count(bom.RELATED_ID)>0 THEN 'Yes' ELSE 'No' END as TEXT) AS IS_ASSY
  FROM RecursiveBOM

INNER JOIN innovator.PART sp ON RecursiveBOM.source_id = sp.ID
INNER JOIN innovator.PART rp ON RecursiveBOM.related_id = rp.ID

--Manufacturer Part
LEFT JOIN innovator.PART_AML aml ON aml.SOURCE_ID = rp.ID
LEFT JOIN innovator.MANUFACTURER_PART mp ON aml.related_ID = mp.ID
LEFT JOIN innovator.MANUFACTURER man on mp.MANUFACTURER = man.ID

--Supplier
LEFT JOIN innovator.VENDOR_PART vp ON mp.ID = vp.RELATED_ID
LEFT JOIN innovator.VENDOR ven ON vp.SOURCE_ID = ven.ID

--Has drawings
LEFT JOIN innovator.CAD_PART cp ON cp.SOURCE_ID = rp.ID

--Is Assembly
LEFT JOIN innovator.PART_BOM bom ON bom.SOURCE_ID = rp.ID

--Don't decend into PCBA
WHERE sp.classification NOT LIKE '%PCBA%'

--Collect the results 
GROUP BY
  rp.ITEM_NUMBER,
  mp.ITEM_NUMBER,
  rp.NAME,
  man.NAME,
  vp.CATALOG_NUMBER,
  ven.NAME,
  cp.RELATED_ID,
  rp.MAKE_BUY

【问题讨论】：

我的猜测是 PART_BOM 与 PART 是 1-n 关系，所以当你加入它时，你会重复行，并使总和不正确
我可能应该稍微重新表述一下这个问题，也许应该改为“我如何获得预期的结果”

标签： sql sql-server-2008

【解决方案1】：

有两种可能的替代解决方案应该更简洁：

1) 使用子选择。这导致查询：

...
SELECT
  rp.MAKE_BUY as MAKE_BUY,
  rp.ITEM_NUMBER AS PN,
  rp.NAME AS NAME,
  sum(RecursiveBOM.quantity) as Total,
  mp.ITEM_NUMBER as MPN,
  man.NAME as MANUFACTURER,
  vp.CATALOG_NUMBER as SPN,
  ven.NAME as SUPPLIER,      
  CAST(CASE WHEN count(cp.RELATED_ID)>0 THEN 'Yes' ELSE 'No' END AS TEXT) AS HAS_CAD,
  (select count(*) from innovator.part_bom bom where bom.source_id = rp.ID) AS IS_ASSY
FROM RecursiveBOM
...

2) 将 CTE 表达式与生成的查询链接在一起：

WITH RecursiveBOM(source_id, related_id, quantity) AS
(
  SELECT pbom.SOURCE_ID, pbom.RELATED_ID, pbom.QUANTITY
    FROM innovator.PART_BOM AS pbom
    WHERE SOURCE_ID = '82FD83221C5946F0A70CADD328BDB6CA'

  UNION ALL

  SELECT pbom.SOURCE_ID, pbom.RELATED_ID, pbom.QUANTITY
    FROM innovator.PART_BOM AS pbom
    INNER JOIN RecursiveBOM rbom ON rbom.RELATED_ID = pbom.SOURCE_ID
), grouping as (
    SELECT
      rp.MAKE_BUY as MAKE_BUY,
      rp.ITEM_NUMBER AS PN,
      rp.NAME AS NAME,
      sum(RecursiveBOM.quantity) as Total,
      mp.ITEM_NUMBER as MPN,
      man.NAME as MANUFACTURER,
      vp.CATALOG_NUMBER as SPN,
      ven.NAME as SUPPLIER,      
      CAST(CASE WHEN count(cp.RELATED_ID)>0 THEN 'Yes' ELSE 'No' END AS TEXT) AS HAS_CAD,
      rp.id related_id
    FROM RecursiveBOM

INNER JOIN innovator.PART sp ON RecursiveBOM.source_id = sp.ID
INNER JOIN innovator.PART rp ON RecursiveBOM.related_id = rp.ID

--Manufacturer Part
LEFT JOIN innovator.PART_AML aml ON aml.SOURCE_ID = rp.ID
LEFT JOIN innovator.MANUFACTURER_PART mp ON aml.related_ID = mp.ID
LEFT JOIN innovator.MANUFACTURER man on mp.MANUFACTURER = man.ID

--Supplier
LEFT JOIN innovator.VENDOR_PART vp ON mp.ID = vp.RELATED_ID
LEFT JOIN innovator.VENDOR ven ON vp.SOURCE_ID = ven.ID

--Has drawings
LEFT JOIN innovator.CAD_PART cp ON cp.SOURCE_ID = rp.ID

--Don't decend into PCBA
WHERE sp.classification NOT LIKE '%PCBA%'

--Collect the results 
GROUP BY
  rp.ITEM_NUMBER,
  mp.ITEM_NUMBER,
  rp.NAME,
  man.NAME,
  vp.CATALOG_NUMBER,
  ven.NAME,
  cp.RELATED_ID,
  rp.MAKE_BUY,
  rp.id
)
select g.make_buy
  , g.pn
  , g.name
  , g.total
  , g.mpn
  , g.manufacturer
  , g.spn
  , g.supplier
  , g.has_cad
  , CASE WHEN count(bom.id) > 0 then 'Yes' ELSE 'No' END as is_assy
from grouping g
LEFT JOIN innovator.PART_BOM bom ON bom.SOURCE_ID = g.related_id
group by g.make_buy
  , g.pn
  , g.name
  , g.total
  , g.mpn
  , g.manufacturer
  , g.spn
  , g.supplier
  , g.has_cad

【讨论】：

【解决方案2】：

我找到了获得我想要的东西的方法，因此我将其添加为答案。但是看起来不优雅，有没有别的办法？

在阅读Insert into a Temp Table in a CTE 的问题时，我意识到我可以通过首先将结果收集到临时表中，然后与 BOM 建立关系来防止行重复。

我相信有更好的方法....

<!-- language: lang-sql -->
create table #TempTable (
  id varchar(32),
  make_buy varchar(100),
  pn varchar(100),
  name varchar(100),
  total int,
  mpn varchar(100),
  manufacturer varchar(100),
  spn varchar(100),
  supplier varchar(100),
  has_cad varchar(100));


WITH RecursiveBOM(source_id, related_id, quantity, prev,next) AS
(
  SELECT pbom.SOURCE_ID, pbom.RELATED_ID, pbom.QUANTITY,0,0
  FROM innovator.PART_BOM AS pbom
  WHERE SOURCE_ID = '82FD83221C5946F0A70CADD328BDB6CA'

  UNION ALL

  SELECT pbom.SOURCE_ID, pbom.RELATED_ID, pbom.QUANTITY, next, prev
  FROM innovator.PART_BOM AS pbom
  INNER JOIN RecursiveBOM rbom ON rbom.RELATED_ID = pbom.SOURCE_ID
)

INSERT INTO #TempTable (id, make_buy, pn, name, total, mpn, manufacturer, spn, supplier, has_cad)

SELECT
  rp.ID,
  rp.MAKE_BUY as MAKE_BUY,
  rp.ITEM_NUMBER AS PN,
  rp.NAME AS NAME,
  sum(RecursiveBOM.quantity) as Total,
  mp.ITEM_NUMBER as MPN,
  man.NAME as MANUFACTURER,
  vp.CATALOG_NUMBER as SPN,
  ven.NAME as SUPPLIER,
  CAST(CASE WHEN count(cp.RELATED_ID)>0 THEN 'Yes' ELSE 'No' END AS TEXT) AS HAS_CAD

FROM RecursiveBOM

INNER JOIN innovator.PART sp ON RecursiveBOM.source_id = sp.ID
INNER JOIN innovator.PART rp ON RecursiveBOM.related_id = rp.ID

--Manufacturer Part
LEFT JOIN innovator.PART_AML aml ON aml.SOURCE_ID = rp.ID
LEFT JOIN innovator.MANUFACTURER_PART mp ON aml.related_ID = mp.ID
LEFT JOIN innovator.MANUFACTURER man on mp.MANUFACTURER = man.ID

--Supplier
LEFT JOIN innovator.VENDOR_PART vp ON mp.ID = vp.RELATED_ID
LEFT JOIN innovator.VENDOR ven ON vp.SOURCE_ID = ven.ID

--Has drawings
LEFT JOIN innovator.CAD_PART cp ON cp.SOURCE_ID = rp.ID

--Don't decend into PCBA
WHERE sp.classification NOT LIKE '%PCBA%'

--Collect the results
GROUP BY
  rp.ID,
  rp.ITEM_NUMBER,
  mp.ITEM_NUMBER,
  rp.NAME,
  man.NAME,
  vp.CATALOG_NUMBER,
  ven.NAME,
  cp.RELATED_ID,
  rp.MAKE_BUY

SELECT
  #TempTable.*,
  CAST(CASE WHEN count(bom.RELATED_ID)>0 THEN 'Yes' ELSE 'No' END as TEXT) AS IS_ASSY

FROM #TempTable

--Is Asseembly
LEFT JOIN innovator.PART_BOM as bom on bom.SOURCE_ID = #TempTable.id

--Collect the results
GROUP BY
  #TempTable.id,
  #TempTable.make_buy,
  #TempTable.pn,
  #TempTable.name,
  #TempTable.total,
  #TempTable.mpn,
  #TempTable.manufacturer,
  #TempTable.spn,
  #TempTable.supplier,
  #TempTable.has_cad

DROP TABLE #TempTable

【讨论】：