Question

我真的不需要将结果缩进，这只是我能想出的最佳标题。非常感激任何的帮助。我花了好几个小时尝试通过CTE这样做，这似乎是要走的路，但我被卡住了。

编辑：我可以通过Component_Job列对下面的示例数据进行排序，但是，在现实世界中，作业编号是随机的，不一定是任何可用的顺序。

我的表格包含以下内容：

Root_Job    Parent_Job  Component_Job  
1           1           1a  
1           1           1b  
1           1           1c  
1           1a          1a1  
1           1a          1a2  
1           1b          1b1  
1           1b          1b2  
2           2           2a  
2           2           2b

我正在尝试创建一个返回以下内容的视图：

Root_Job    Parent_Job  Component_Job
1           1           1a
1           1a          1a1
1           1a          1a2
1           1           1b
1           1b          1b1
1           1b          1b2
1           1           1c
2           2           2a
2           2           2b

只是为了澄清我想要达到的回报顺序：

最后，我一直尝试的CTE，但对我无能为力：

with BOM (Root_job, parent_job, component_Job)
as
(
-- Anchor member definition
    SELECT e.Root_Job, e.Parent_Job, e.Component_Job
    FROM Bill_Of_Jobs AS e
    WHERE Root_Job = Parent_Job
    UNION ALL
-- Recursive member definition
    SELECT e.Root_Job, e.Parent_Job, e.Component_Job
    FROM Bill_Of_Jobs AS e
    INNER JOIN bill_of_Jobs AS d
    ON e.parent_Job = d.Component_Job
)
-- Statement that executes the CTE
SELECT * from BOM

Answer 1

SELECT *
FROM BOM
ORDER BY LEFT(Component_Job+'000',3)

Answer 2

这里可能有用：

declare @Jobs as Table ( ParentJob VarChar(10), ComponentJob VarChar(10) );
insert into @Jobs ( ParentJob, ComponentJob ) values
  ( '1', '1a' ), ( '1', '1b' ), ( '1', '1c' ),
  ( '1a', '1a1' ), ( '1a', '1a2' ), ( '1b', '1b1' ), ( '1b', '1b2' ),
  ( '2', '2a' ), ( '2', '2b' );

select * from @Jobs;

with Roots as (
  -- Find and fudge the root jobs.
  --   Usually they are represented as children without parents, but here they are implied by the presence of children.
  select distinct 1 as Depth, ParentJob as RootJob, Cast( ParentJob as VarChar(1024) ) as Path, ParentJob, ParentJob as ComponentJob
    from @Jobs as J
    where not exists ( select 42 from @Jobs where ComponentJob = J.ParentJob ) ),
  BoM as (
  -- Anchor the indented BoM at the roots.
  select Depth, RootJob, Path, ParentJob, ComponentJob
    from Roots
  union all
  -- Add the components one level at a time.
  select BoM.Depth + 1, BoM.RootJob, Cast( BoM.Path + '»' + J.ComponentJob as VarChar(1024) ), J.ParentJob, J.ComponentJob
    from BoM inner join
      @Jobs as J on J.ParentJob = BoM.ComponentJob )
  -- Show the result with indentation.
  select *, Space( Depth * 2 ) + ComponentJob as IndentedJob
    from BoM
    order by ComponentJob
    option ( MaxRecursion 0 );

在现实世界中，很少有能够轻松排序的东西。数字项（1,1.1,1.1.1,1.2）的技巧是创建一个Path，其每个值都填充为零填充到固定长度，例如： 0001，0001»0001，0001»0001»0001，0001»0002，以便在按字母顺序排列时排序正确。您的数据可能会有所不同。

以“缩进”顺序返回BOM结果

2 个答案: