在Oracle SQL中循环

Question

我想将ID_last列添加到下面左侧的表中。 示例表由6行和8行的两个序列组成。 我需要处理许多序列，每个序列的行数/每个序列的循环数未知/无限。 当ID不在ID_new中时，序列开始；当ID_new不在ID中时，序列停止。 我想知道序列中所有ID的最后一个ID_new（如右表所示）。

我该如何使用SQL（oracle）？

非常感谢帮助！

Answer 1

由于可以使用此查询确定任何顺序的最后一个ID：

select id_new from YourData yd
 where not exists (select 1 from YourData yd2
                    where yd2.id = yd.id_new);

我们可以将问题视为基于已标识ID_Last值的层次查询，并将根ID作为ID_Last返回：

select id
     , id_new
     , CONNECT_BY_ROOT id_new ID_Last
  from YourData yd
  connect by NOCYCLE id_new = prior id
 start with not exists (select 1 from YourData yd2
                         where yd2.id = yd.id_new)

或者，可以使用上面的第一个查询（带有其他列）作为锚点查询，将其编写为递归查询。 为了避免循环，我们必须添加一列来跟踪已被访问的节点，并在查询的递归部分中对其进行检查：

With Recur(id, id_new, id_last, nodes) as (
  select id
       , id_new
       , id_new
       , ':'||id||':'
    from YourData yd1
   where not exists (select 1 from YourData yd2
                      where yd2.id = yd1.id_new)
   union all
  select yd.id
       , yd.id_new
       , r.id_last
       , r.nodes||yd.id||':'
    from YourData yd
    join Recur r
      on r.id = yd.id_new
     and r.nodes not like '%:'||yd.id||':%' -- Avoid cycles
)
select id, id_new, id_last
  from Recur
 order by id_last
     , nodes desc;

Answer 2

您没有提供表名，所以我假设它的名称为table1 。

这是一个查询，用于查找层次结构中每个节点的叶节点。 请注意，如果您的任何节点有多个子节点，您将获得一些重复。

-- sample data
with table1 as (select 11 as id, 12 as id_new from dual
    union all select 12, 5 from dual
    union all select 5, 18 from dual
    union all select 18, 17 from dual
    union all select 17, 28 from dual
    union all select 28, 13 from dual
    union all select 25, 31 from dual
    union all select 31, 22 from dual
    union all select 22, 41 from dual
    union all select 41, 33 from dual
    union all select 33, 39 from dual
    union all select 39, 30 from dual
    union all select 30, 45 from dual
    union all select 45, 24 from dual)
-- query
select regexp_substr(id_path, '[^>]+', 1, 1) as root, 
  id_new as id_last
from (select CONNECT_BY_ISLEAF isleaf, sys_connect_by_path(id, '>') as id_path, id_new
      from table1
      connect by prior id_new = id)
where isleaf = 1
;

我认为这应该作为更新语句，但是我尚未对其进行测试。

merge into table1 t
  using (select regexp_substr(id_path, '[^>]+', 1, 1) root, id_new as id_last
        from (select CONNECT_BY_ISLEAF isleaf, sys_connect_by_path(r.id, '>') as id_path, r.id_new
              from table1 r
              connect by prior r.id_new = r.id)
        where isleaf = 1) u
  on (t.id = u.root)
when matched then update
  set t.id_last = u.id_last;

Answer 3

如果您具有Oracle 11gR2或更高版本，则可以使用像这样的递归查询（假设您的表称为tbl ）：

with tbl2(id, id_new, id_last, lvl) as (
    select     id, id_new, id_new, 1 
    from       tbl
    union all
    select     tbl2.id, tbl2.id_new, tbl.id_new, tbl2.lvl+1 
    from       tbl2
    inner join tbl 
            on tbl.id = tbl2.id_last
)
select unique id, id_new, 
       first_value(id_last) over (partition by id, id_new order by lvl desc) id_last 
from   tbl2

这不涉及循环（我没有对此发表评论的回复）。