Ddeleting the Second Row tag in XML output

Question

Hi All,
When I run the following query:
select dbms_xmlgen.getxml('select wo.work_order_no,
cursor(select regacc.work_order_no, regacc.work_order_task_no, regacc.regulatory_account_no
from sa_work_orde_reg regacc
where regacc.work_order_no = wo.work_order_no) Reg_Test
from sa_work_order wo') xml
from dual;
I get the following output:
<?xml version="1.0"?>
<ROWSET>
<ROW>
<WORK_ORDER_NO>1000170</WORK_ORDER_NO>
<REG_TEST>
<REG_TEST_ROW>
<WORK_ORDER_NO>1000170</WORK_ORDER_NO>
<WORK_ORDER_TASK_NO>01</WORK_ORDER_TASK_NO>
<REGULATORY_ACCOUNT_NO>100.2.1</REGULATORY_ACCOUNT_NO>
</REG_TEST_ROW>
</REG_TEST>
</ROW>
</ROWSET>
Now what I need is to delete the second row tag <REG_TEST_ROW>. Is there some function or some other ways to handle this. I will be using this query in DBMS_XMLGEN.newcontext API inside my procedure.
Thanks for your help in advance.
Edited by: user33333 on Jun 29, 2010 1:11 AM
Edited by: user33333 on Jun 29, 2010 1:19 AM

Answer 1

Hi,
Here are some options...
1) Using a scalar subquery instead of the CURSOR constructor (provided it's always meant to return a single row) :
SELECT dbms_xmlgen.getxml(
'SELECT wo.work_order_no,
        ( SELECT XMLForest( regacc.work_order_no,
                            regacc.work_order_task_no,
                            regacc.regulatory_account_no )
          FROM sa_work_orde_reg regacc
          WHERE regacc.work_order_no = wo.work_order_no ) reg_test
FROM sa_work_order wo'
) xml
FROM dual;2) Embedding an XSL transformation (yes, again!) to copy all nodes except REG_TEST_ROW :
<?xml version="1.0"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="*[local-name()!='REG_TEST_ROW']">
<xsl:copy select=".">
  <xsl:apply-templates/>
</xsl:copy>
</xsl:template>
<xsl:template match="*[local-name()='REG_TEST_ROW']">
<xsl:copy-of select="./*"/>
</xsl:template>
</xsl:stylesheet>3) Using a simple replace on the serialized output.