SQL 服务器 : BOM Recursion from the bottom up

Question

我试图自下而上地总结 BOM 成本。我需要能够在 BOM 的特定级别确定该级别的成本是多少，因为所有成本都从下面的级别累积。

在下面的示例中，作业 1000 的成本应该是下面所有作业的所有成本的总和以及作业 1000 的成本。1000-1 应该是 1000-1 + 1000-1A 的总和，1000-2 将仅包含 1000-2 的成本，因为没有与该作业相关的组件，等等...

(注意:工作编号在现实世界中是随机的，无法可靠地排序。)

1000              
   1000-1
      1000-1a
      1000-1B
         1000-1B1 
   1000-2
   1000-3
      1000-3A
      1000-3B
         1000-3B-1
   1000-4

Bill_Of_Jobs 定义装配/BOM 结构以及包含每个作业的成本核算信息的 Job 表。

在下面的示例中，我希望返回:

1000      = $150
1000-1    = $140
1000-1A   = $ 30
1000-1B   = $ 90
1000-1B-1 = $ 50

SQL Fiddle Example

CREATE TABLE [Bill_Of_Jobs]
(
    [Parent_Job] varchar(10) NOT NULL,
    [Component_Job] varchar(10) NOT NULL,
    [Root_Job] varchar(10) NULL,
)

Insert into Bill_Of_Jobs (Parent_Job, Component_Job, Root_Job)
Values ('1000', '1000-1', '1000'),
       ('1000-1', '1000-1A', '1000'),
       ('1000-1', '1000-1B', '1000'),
       ('1000-1B', '1000-1B-1', '1000')

Create Table Job 
(
    Job varchar(10), 
    Top_Lvl_Job varchar(10), 
    [Type] varchar(10), 
    Act_Material money
)

Insert into Job (Job, Top_Lvl_Job, [Type], Act_Material)
Values ('1000', '1000', 'Assembly', 10.00),
       ('1000-1', '1000', 'Assembly', 20.00),
       ('1000-1A', '1000', 'Regular', 30.00),
       ('1000-1B', '1000', 'Assembly', 40.00),
       ('1000-1B-1', '1000', 'Regular', 50.00)

下面的查询与我所能得到的一样接近。它没有正确求和，而是自上而下与自下而上求和。非常感谢任何帮助。

WITH roots AS 
(
    SELECT DISTINCT 
        1 AS [Level], 
        BOJ.parent_job AS RootJob, 
        Cast(BOJ.parent_job AS VARCHAR(1024)) AS Path, 
        BOJ.parent_job, 
        BOJ.parent_job AS ComponentJob, 
        job.act_material 
    FROM   
        bill_of_jobs AS BOJ 
    INNER JOIN 
        job  ON BOJ.parent_job = job.job 
    WHERE  
        (NOT EXISTS (SELECT 'z' AS Expr1 
                     FROM bill_of_jobs 
                     WHERE (component_job = BOJ.parent_job)
                    )
        )
), 
bom AS 
(
    SELECT 
        [level], 
        rootjob, 
        path, 
        parent_job, 
        componentjob, 
        act_material 
    FROM   
        roots 

    UNION ALL 

    SELECT 
        bom.[level] + 1, 
        bom.rootjob, 
        Cast(bom.path + '»' + BOJ2.component_job AS VARCHAR(1024)), 
        BOJ2.parent_job, 
        BOJ2.component_job, 
        bom.act_material + J.act_material 
    FROM   
        bom 
    INNER JOIN 
        bill_of_jobs AS BOJ2 ON BOJ2.parent_job = bom.componentjob 
    INNER JOIN 
        job AS J ON BOJ2.component_job = J.job
) 
SELECT 
    componentjob AS Component_Job, 
    [path], 
    Space( [level] * 2 ) + componentjob AS IndentedBOM, 
    Dense_rank() OVER (partition BY rootjob ORDER BY path) AS View_Order, 
    act_material 
FROM   
    bom   

Answer 1

您可以使用此递归 CTE:

;WITH BottomUp AS ( 
    SELECT Component_Job, Parent_Job, j.Act_Material, 1 AS level
    FROM Bill_Of_Jobs AS b
    INNER JOIN Job AS j ON b.Component_Job = j.Job

    UNION ALL

    SELECT c.Component_Job, b.Parent_Job, j.Act_Material, level = c.level + 1
    FROM Bill_Of_Jobs AS b  
    INNER JOIN BottomUp c ON c.Parent_Job = b.Component_Job
    INNER JOIN Job AS j ON c.Component_Job = j.Job   
)
SELECT *
FROM BottomUp

获取每个 Component_Job 的所有祖先:

Component_Job   Parent_Job  Act_Material    level
-------------------------------------------------
1000-1          1000        20,00           1  
1000-1A         1000-1      30,00           1
1000-1B         1000-1      40,00           1
1000-1B-1       1000-1B     50,00           1
1000-1B-1       1000-1      50,00           2
1000-1B-1       1000        50,00           3
1000-1B         1000        40,00           2
1000-1A         1000        30,00           2

如果您UNION 上面的结果集带有叶节点:

;WITH BottomUp AS (
  ... above query here
), BottomUpWithLeafNodes AS (
   SELECT Component_Job, Parent_Job, Act_Material, level
   FROM BottomUp 

   UNION 

   SELECT Job AS Component_Job, Job AS Parent_Job, Act_Material, 0 AS level
   FROM Job
   WHERE Job NOT IN (SELECT Parent_Job FROM Bill_Of_Jobs)
)
SELECT *
FROM BottomUpWithLeafNodes

然后你有一些你可以 GROUP BY Parent_Job 列。您只需为每组非叶节点的父节点添加 Act_Material 值即可获得所需的结果:

;WITH BottomUp AS (
     ... above query here
), BottomUpWithLeafNodes AS (
     ... above query here
)
SELECT Parent_Job AS Job, 
       SUM(Act_Material) + CASE 
                              WHEN SUM(level) <> 0 THEN (SELECT Act_Material FROM Job WHERE Job = b.Parent_Job)
                             ELSE 0
                           END AS Act_Material 
FROM BottomUpWithLeafNodes AS b
GROUP BY Parent_Job

输出:

Parent_Job  Act_Material
------------------------
1000        150,00
1000-1      140,00
1000-1A     30,00
1000-1B     90,00
1000-1B-1   50,00

SQL Fiddle Demo