Select XMLElement - want some whitespace or newlines!

Question

Hi -
I'm using XML SQL psuedo functions to make an XML document. It's making valid XML, but it's hideous to read. The SQL select is actually a select ... into in a PL/SQL function. Is there any way to force some newlines or spaces after the end tags, so the message would be more readable in a text editor?
EXAMPLE:
SELECT XMLElement("OBSERVATION-DATA",
     XMLAttributes('http://www.w3.org/2001/XMLSchema-instance' AS "xmlns:xsi",
          'C:\Documents and Settings\rmercer\My Documents\Rick\projects\LAPHIN\Maps\canonical_v2\canonical_lab_v3.xsd' AS "xsi:noNamespaceSchemaLocation"),
XMLForest(
     XMLForest(
          c_SOURCEEXT     as "SOURCE-IDENT",
          c_SOURCEROOT     as "SOURCE-ROOT",
          c_SOURCEEXT     as "SOURCE-FACILITY-IDENT",
          c_SOURCEROOT     as "SOURCE-FACILITY-ROOT",
          ) as "SOURCE-INFORMATION", -- end of source info forest
     XMLForest(
          XMLConcat( XMLElement("PERSON-ID-NUMBER", MPERSON.PERSON_UID),
     ) -- end of main forest
     ) into v_OBSXML
FROM nbs_msgine.msgin_person MPERSON,
GIVES A REALLY LONG LINE OF OUTPUT:
<SOURCE-INFORMATION><SOURCE-IDENT>LABCORP</SOURCE-IDENT><SOURCE-ROOT>999.5.1.2.1</SOURCE-ROOT><SOURCE-FACILITY-IDENT>LABCORP</SOURCE-FACILITY-IDENT><SOURCE-FACILITY-ROOT>999.5.1.2.1</SOURCE-FACILITY-ROOT><RECVR-IDENT>LADOH</RECVR-IDENT><RECVR-ROOT>999.5.1.1.5</RECVR-ROOT><RECVR-FACILITY-IDENT>LADOH</RECVR-FACILITY-IDENT><RECVR-FACILITY-
Thanks!
Andy Russo

Answer 1

The SQL/XML functions do not pretty-print for performance reasons. Over a large document the amount of additional white space generated by pretty printing can be quite expensive.
There is a very simple way of forcing a pretting print
As can be seen in the example below the output is not pretty printed. It's technically correct but difficult to read.
SQL> select xmlElement
2 (
3 "Country",
4 xmlElement("Name",c.COUNTRY_NAME),
5 xmlAgg
6 (
7 xmlElement
8 (
9 "Department",
10 xmlElement("Name", d.DEPARTMENT_NAME),
11 ( select xmlAgg
12 (
13 xmlElement
14 (
15 "Employee",
16 xmlForest(e.FIRST_NAME as "FirstName", e.LAST_NAME as "LastName")
17 )
18 )
19 from HR.EMPLOYEES e
20 where e.DEPARTMENT_ID = d.DEPARTMENT_ID
21 )
22 )
23 )
24 ) as XML
25 from HR.COUNTRIES c, HR.DEPARTMENTS d, HR.LOCATIONS l
26 where l.COUNTRY_ID = c.COUNTRY_ID
27 and d.LOCATION_ID = l.lOCATION_ID
28 and c.COUNTRY_NAME = 'Canada'
29 group by c.COUNTRY_NAME
30 /
<Country><Name>Canada</Name><Department><Name>Marketing</Name><Employee><FirstNa
me>Michael</FirstName><LastName>Hartstein</LastName></Employee><Employee><FirstN
ame>Pat</FirstName><LastName>Fay</LastName></Employee></Department></Country>
In order to get pretty printed output we need to force a print & parse. It should be remembered that this is fairly expensive and so should only be done when debugging out etc. It can be done by adding .extract('/*') to the output of the outermost SQL/XML operator.
Eg
SQL> select xmlElement
2 (
3 "Country",
4 xmlElement("Name",c.COUNTRY_NAME),
5 xmlAgg
6 (
7 xmlElement
8 (
9 "Department",
10 xmlElement("Name", d.DEPARTMENT_NAME),
11 ( select xmlAgg
12 (
13 xmlElement
14 (
15 "Employee",
16 xmlForest(e.FIRST_NAME as "FirstName", e.LAST_NAME as "LastName")
17 )
18 )
19 from HR.EMPLOYEES e
20 where e.DEPARTMENT_ID = d.DEPARTMENT_ID
21 )
22 )
23 )
24 ).extract('/*') as XML
25 from HR.COUNTRIES c, HR.DEPARTMENTS d, HR.LOCATIONS l
26 where l.COUNTRY_ID = c.COUNTRY_ID
27 and d.LOCATION_ID = l.lOCATION_ID
28 and c.COUNTRY_NAME = 'Canada'
29 group by c.COUNTRY_NAME
30 /
<Country>
<Name>Canada</Name>
<Department>
<Name>Marketing</Name>
<Employee>
<FirstName>Michael</FirstName>
<LastName>Hartstein</LastName>
</Employee>
<Employee>
<FirstName>Pat</FirstName>
<LastName>Fay</LastName>
</Employee>
</Department>
</Country>