Nested group function without group xmlagg

Question

I am getting nested group function without group by xmlagg when using the xmlagg function inside another xmlagg function. Find the table structure and sample data here,
CREATE TABLE "TEST_TABLE"
   ("KEY" NUMBER(20,0),
    "NAME" VARCHAR2(50 ),
    "DESCRIPTION" VARCHAR2(100 )
   Insert into TEST_TABLE (KEY,NAME,DESCRIPTION) values (1,'sam','desc1');
   Insert into TEST_TABLE (KEY,NAME,DESCRIPTION) values (2,'max','desc2');
   Insert into TEST_TABLE (KEY,NAME,DESCRIPTION) values (3,'peter',null);
   Insert into TEST_TABLE (KEY,NAME,DESCRIPTION) values (4,'andrew',null);
select
        XMLSerialize(document
        xmlelement("root",
         xmlagg(
           xmlelement("emp"          
           , xmlforest(Key as "ID")          
           , xmlforest(name as "ename")
           , xmlelement("Descriptions", 
           xmlagg(
              xmlforest(description as "Desc")
       ) as clob indent
       ) as t   
      from test_table;Then i removed the xmlagg function from the above select query and used xmlelement instead
  select
        XMLSerialize(document
        xmlelement("root",
         xmlagg(
           xmlelement("emp"          
           , xmlforest(Key as "ID")          
           , xmlforest(name as "ename")
           , xmlelement("Descriptions",            
              xmlforest(description as "Desc")
       ) as clob indent
       ) as t   
      from test_table;This is working fine, but xml created with empty elements for Descriptions element for key 3 and 4 which has null values. I need don't need Descriptions element in the xml when it has null value. Please help me to resolve this.

Answer 1

You can do it with a correlated subquery :
SQL> select xmlserialize(document
  2           xmlelement("root",
  3             xmlagg(
  4               xmlelement("emp"
  5               , xmlforest(
  6                   t.key as "ID"
  7                 , t.name as "ename"
  8                 , (
  9                     select xmlagg(
10                              xmlelement("Desc", d.description)
11                              order by d.description -- if necessary
12                            )
13                     from test_desc d
14                     where d.key = t.key
15                   ) as "Descriptions"
16                 )
17               )
18             )
19           ) as clob indent
20         )
21  from test_table t;
XMLSERIALIZE(DOCUMENTXMLELEMEN
<root>
  <emp>
    <ID>1</ID>
    <ename>sam</ename>
    <Descriptions>
      <Desc>desc1_1</Desc>
      <Desc>desc1_2</Desc>
      <Desc>desc1_3</Desc>
    </Descriptions>
  </emp>
  <emp>
    <ID>2</ID>
    <ename>max</ename>
    <Descriptions>
      <Desc>desc2_1</Desc>
      <Desc>desc2_2</Desc>
      <Desc>desc2_3</Desc>
    </Descriptions>
  </emp>
  <emp>
    <ID>3</ID>
    <ename>peter</ename>
  </emp>
  <emp>
    <ID>4</ID>
    <ename>andrew</ename>
  </emp>
</root>
Or an OUTER JOIN + GROUP-BY :
select xmlserialize(document
         xmlelement("root",
           xmlagg(
             xmlelement("emp"          
             , xmlforest(
                 t.key as "ID"
               , t.name as "ename"
               , xmlagg(
                   xmlforest(d.description as "Desc")
                   order by d.description -- if necessary
                 ) as "Descriptions"
         ) as clob indent
from test_table t
     left outer join test_desc d on d.key = t.key
group by t.key
       , t.name
;Edited by: odie_63 on 11 juil. 2012 14:54 - added 2nd option