Re: Question about XML and Oracle.

From: Gerard H. Pille <ghp_at_skynet.be>
Date: Thu, 02 Dec 2010 21:06:38 +0100
Message-ID: <4cf7fc4a$0$14255$ba620e4c_at_news.skynet.be>

John Peterson wrote:
>
> "Gerard H. Pille" <ghp_at_skynet.be> wrote in message news:4ce6b41f$0$14255$ba620e4c_at_news.skynet.be...
>>
>> SELECT XMLROOT ( XMLElement("OuterElement",
>> "<AllMyQueryResultsXML/>"
>> ), VERSION '1.0', STANDALONE YES)
>> AS "XMLROOT" FROM DUAL;
>>
>>
>
> Thanks Gerald -- but I can't seem to figure out how to structure the query inside of the outer
> query. Maxim's reply seems to be (sort of) working for me, so I'll probably go that route.

Pourtant, c'est simple comme bonjour:

XMLROOT

---

<?xml version="1.0" standalone="yes"?>
<OuterElement>

   <Topic>

     <TOPICID>1</TOPICID>
     <TOPICDESC>This is the first topic</TOPICDESC>
     <AREA>0</AREA>

   </Topic>
   <Topic>

     <TOPICID>2</TOPICID>
     <TOPICDESC>This is the second topic</TOPICDESC>
     <AREA>0</AREA>

   </Topic>
   <Topic>

     <TOPICID>3</TOPICID>
     <TOPICDESC>This is the third topic</TOPICDESC>
     <AREA>1</AREA>

   </Topic>
   <Topic>

     <TOPICID>4</TOPICID>
     <TOPICDESC>This is the fourth topic</TOPICDESC>
     <AREA>1</AREA>

   </Topic>
</OuterElement>

SQL> l

   1 SELECT XMLROOT (
   2 XMLElement(

   3      "OuterElement",
   4      xmlagg(
   5        XMLElement(
   6          "Topic",
   7          xmlelement("TOPICID",TOPICID),
   8          xmlelement("TOPICDESC",TOPICDESC),
   9          xmlelement("AREA",AREA)
  10          )
  11        )
  12      ), VERSION '1.0', STANDALONE YES)

  13* AS "XMLROOT" FROM faq.topic Received on Thu Dec 02 2010 - 14:06:38 CST