从XML到Oracle PL/SQL环境中的路径列表 [英] From XML to list of paths in Oracle PL/SQL environment

问题描述

请假定您有一个XML文件(例如，存储在还具有CLOB列的Oracle表中):

Please suppose you have a XML file (stored, for example, in an Oracle table which has also a CLOB column):

<ALFA>
  <BETA>0123</BETA>
  <GAMMA>2345</GAMMA>
  <DELTA>
     <EPSILON>3</EPSILON>
  </DELTA>
</ALFA>

如何在输出中生成所有可能路径的列表?

How can I produce, in output, the list of all possible paths?

/ALFA/BETA/text()
/ALFA/GAMMA/text()
/ALFA/DELTA/EPSILON/text()

我的需求如下:我必须从长XML中提取许多信息，并且必须将XMLEXTRACT与所有可能的路径一起使用，因此我想知道是否有可能自动将它们"dbms_output.put_line"方式.

My need is the following: I have to EXTRACT many information from a long XML and I have to use XMLEXTRACT with all possible paths, so I would like to know if is it possible to "dbms_output.put_line" them in an automatic way.

我需要一个独立于标签名称的解决方案.

I need a solution which is independent from the name of the tags.

请假定XML格式正确.

Please suppose that the XML is well-formed.

在此先感谢您的帮助.

• 在第二种情况下:

如果尚未安装Oracle Java Extension，如何处理并收到以下错误?

How can I proceed if Oracle Java Extension has not been installed, and I receive the following error?

ORA-19112: error raised during evaluation:  
ORA-06550: line 1, column 13:
PLS-00201: identifier 'SYS.DBMS_XQUERYINT' must be declared
ORA-06550: line 1, column 7:
PL/SQL: Statement ignored

请假设我不是DBA，并且DBA不授权Oracle Java Extension安装.

Please suppose that I am not DBA, and DBA don't authorize Oracle Java Extension installation.

推荐答案

您可以使用 XMLTable 以使用XQuery生成路径列表.

You can use XMLTable to produce list of paths with XQuery.

例如

( SQLFiddle )

with params as (
  select 
    xmltype('
      <ALFA>
        <BETA>0123</BETA>
        <GAMMA>2345</GAMMA>
        <DELTA>
           <EPSILON>3</EPSILON>
        </DELTA>
      </ALFA>
    ') p_xml
  from dual  
)    
select
  path_name || '/text()'
from
  XMLTable(
    '
      for $i in $doc/descendant-or-self::*
        return <element_path> {$i/string-join(ancestor-or-self::*/name(.), ''/'')} </element_path>
    '
    passing (select p_xml from params) as "doc"
    columns path_name varchar2(4000) path '//element_path'
  )

但这是一种错误的方式，至少是因为它没有达到预期的效果.

but it's a wrong way at least because it's not effective as it can.

只需使用相同的XQuery提取所有值: ( SQLFiddle )

Just extract all values with same XQuery: (SQLFiddle)

with params as (
  select 
    xmltype('
      <ALFA>
        <BETA>0123</BETA>
        <GAMMA>2345</GAMMA>
        <DELTA>
           <EPSILON>3</EPSILON>
        </DELTA>
      </ALFA>
    ') p_xml
  from dual  
)    
select
  element_path, element_text
from
  XMLTable(
    '              
      for $i in $doc/descendant-or-self::*
        return <element>
                 <element_path> {$i/string-join(ancestor-or-self::*/name(.), ''/'')} </element_path>
                 <element_content> {$i/text()}</element_content>
               </element>  
    '
    passing (select p_xml from params) as "doc"
    columns 
      element_path   varchar2(4000) path '//element_path',
      element_text   varchar2(4000) path '//element_content'
  )

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