在oracle中从CONNECT-BY查询中删除重复的子树 [英] Removing duplicate subtrees from CONNECT-BY query in oracle

问题描述

我有一个格式为

CREATE TABLE tree_hierarchy (
  id        NUMBER (20)
 ,parent_id NUMBER (20)
);


INSERT INTO tree_hierarchy (id, parent_id) VALUES (2, 1);
INSERT INTO tree_hierarchy (id, parent_id) VALUES (4, 2);
INSERT INTO tree_hierarchy (id, parent_id) VALUES (9, 4);

当我运行查询时:-

SELECT id,parent_id,
  CONNECT_BY_ISLEAF leaf,
  LEVEL,
  SYS_CONNECT_BY_PATH(id, '/') Path,
  SYS_CONNECT_BY_PATH(parent_id, '/') Parent_Path
FROM tree_hierarchy
WHERE CONNECT_BY_ISLEAF<>0
  CONNECT BY PRIOR id = PARENT_id
ORDER SIBLINGS BY ID;

我得到的结果是这样的:-

Result I am Getting is like this:-

"ID"    "PARENT_ID" "LEAF"  "LEVEL" "PATH"  "PARENT_PATH"

9            4         1       3    "/2/4/9"  "/1/2/4"

9            4         1       2     "/4/9"     "/2/4"

9            4         1       1      "/9"      "/4"

但是我需要一个只让我得到这个的Oracle Sql查询

But I need an Oracle Sql Query That gets me only this

"ID"    "PARENT_ID" "LEAF"  "LEVEL" "PATH"  "PARENT_PATH"

9            4         1       3    "/2/4/9"  "/1/2/4"

这是一个简单的示例，我以这种方式拥有1000多个记录.当我运行上述查询时，它会生成许多重复项.任何人都可以给我一个泛型查询，该查询将给出从叶到根的完整路径而没有重复.感谢您的提前帮助

This is a simpler example I have more that 1000 records in such fashion.When I run the above query,It is generating many duplicates.Can any one give me a generic query that will give complete path from leaf to root with out duplicates.Thanks for the help in advance

推荐答案

必须始终知道有限层次结构中的根节点. 根据定义: http://en.wikipedia.org/wiki/Tree_structure 根节点是没有父节点的节点. 要检查给定节点是否为根节点，请使用"parent_id"并在表中检查是否存在具有该ID的记录. 该查询可能看起来像这样:

The root node in finite hierarchy must be always known. According to the definition: http://en.wikipedia.org/wiki/Tree_structure the root node is a node that has no parents. To check if a given node is a root node, take "parent_id" and check in the table if exists a record with this id. The query might look like this:

SELECT id,parent_id,
  CONNECT_BY_ISLEAF leaf,
  LEVEL,
  SYS_CONNECT_BY_PATH(id, '/') Path,
  SYS_CONNECT_BY_PATH(parent_id, '/') Parent_Path
FROM tree_hierarchy th
WHERE CONNECT_BY_ISLEAF<>0
  CONNECT BY PRIOR id = PARENT_id
START WITH not exists (
      select 1 from tree_hierarchy th1 
      where th1.id = th.parent_id
  )
ORDER SIBLINGS BY ID;

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