在 XML 中显示层次结构

Question

我有以下 table:

员工 (id int,name varchar,managerid int)

   ID   NAME   MANAGERID
    1    A       2
    2    B       4
    3    C       4
    4    D      NULL

期望输出：

<Node name="D" id="4">
   <Node name="B" id="2">
    <Node name="A" id="1">
    </Node>
   </Node>  
  <Node name="C" id="3">
  </Node>
</Node>

现在，我知道这不过是深度优先搜索，所以我完成了以下操作：

WITH t1(id,name,managerid) AS (
  -- Anchor member.
  SELECT id,
         name,
         managerid
  FROM   employee
    WHERE  managerid IS NULL
  UNION ALL
  -- Recursive member.
  SELECT t2.id,
         t2.name,
         t2.managerid
  FROM   employee t2, t1
  WHERE  t2.managerid = t1.id
)
SEARCH DEPTH FIRST BY id SET order1
SELECT id,
       name,
       managerid
FROM   t1
ORDER BY order1\

而上述查询的输出是：

   ID  NAME MANAGERID
    4   D    NULL
    2   B    4
    1   A    2
    3   C    4

现在，我不知道如何将此输出转换为之前显示的 xml 版本。

我知道有 XMLElement、XMLAGG 等函数，但我不知道如何在这种分层查询中使用它。

注意：目前，我在 Oracle 中执行此操作，但如果另一个 RDBMS 有更简单的方法来解决此问题，那么我完全赞成。

Answer 1

您可以使用 dbms_xmlgen.newcontextFromHierarchy 执行此操作，如下所示：

select dbms_xmlgen.getxmltype(dbms_xmlgen.newcontextFromHierarchy('
WITH sample_data AS (SELECT 1 ID, ''A'' NAME, 2 managerid FROM dual UNION ALL
                     SELECT 2 ID, ''B'' NAME, 4 managerid FROM dual UNION ALL
                     SELECT 3 ID, ''C'' NAME, 4 managerid FROM dual UNION ALL
                     SELECT 4 ID, ''D'' NAME, NULL FROM dual)
select level 
       , xmlelement("node" 
       , XMLAttributes(name as "name",
                       id as "id")
       ) 
FROM   sample_data
CONNECT BY PRIOR ID = managerid
START WITH managerid IS NULL
  ')) from dual;

哪个returns:

<?xml version="1.0"?>
<node name="D" id="4">
  <node name="B" id="2">
    <node name="A" id="1"/>
  </node>
  <node name="C" id="3"/>
</node>

您需要将 dbms_xmlgen.newcontextFromHierarchy() 调用中的查询替换为您的实际查询。

Answer 2

如果您模拟 connect by 的 level 列，则可以使用 DBMS_XMLGEN.newcontextfromhierarchy 和 CTE 完成此操作：

SELECT DBMS_XMLGEN.getXML(DBMS_XMLGEN.newcontextfromhierarchy('
with employee as (
  select 1 id, ''A'' name, 2 managerid from dual union all
  select 2 id, ''B'' name, 4 managerid from dual union all
  select 3 id, ''C'' name, 4 managerid from dual union all
  select 4 id, ''D'' name, null managerid from dual 
)
, t1(lvl,id,name,managerid) AS (
  -- Anchor member.
  SELECT 1 as lvl,
         id, 
         name,
         managerid
  FROM   employee
    WHERE  managerid IS NULL
  UNION ALL
  -- Recursive member.
  SELECT t1.lvl+1 as lvl, 
         t2.id, 
         t2.name,
         t2.managerid
  FROM   employee t2, t1
  WHERE  t2.managerid = t1.id
)
SEARCH DEPTH FIRST BY id SET order1
SELECT lvl, xmlelement("Node", xmlattributes(name AS "name", id AS "id"))
FROM   t1
ORDER BY order1
')) 
FROM dual

输出

<?xml version="1.0"?>
<Node name="D" id="4">
  <Node name="B" id="2">
    <Node name="A" id="1"/>
  </Node>
  <Node name="C" id="3"/>
</Node>

在 XML 中显示层次结构

Display the hierarchy in XML

xml

sql

oracle

hierarchical-data