How to get the column name and table name from xml file

Question

I have one XML file, I generated xsd file from that xml file but the problem is i dont know table name and column name. So my question is how can I retrieve the data from that xml file?

Answer 1

Here's an example using binary XML storage (instead of Object-Relational storage as described in the article).
begin
  dbms_xmlschema.registerSchema(
    schemaURL       => 'my_schema.xsd'
  , schemaDoc       => xmltype(bfilename('TEST_DIR','my_schema.xsd'), nls_charset_id('AL32UTF8'))
  , local           => true
  , genTypes        => false
  , genTables       => true
  , enableHierarchy => dbms_xmlschema.ENABLE_HIERARCHY_CONTENTS
  , options         => dbms_xmlschema.REGISTER_BINARYXML
end;
genTables => true : means that a default schema-based XMLType table will be created during registration.
enableHierarchy => dbms_xmlschema.ENABLE_HIERARCHY_CONTENTS : indicates that a repository resource conforming to the schema will be automatically stored in the default table.
If the schema is not annotated, the name of the default table is system-generated but derived from the root element name :
SQL> select table_name
  2  from user_xml_tables
  3  where xmlschema = 'my_schema.xsd'
  4  and element_name = 'employee';
TABLE_NAME
employee1121_TAB
(warning : the name is case-sensitive)
To annotate the schema and control the naming, modify the content to :
<?xml version="1.0" encoding="UTF-8" ?>
<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:xdb="http://xmlns.oracle.com/xdb">
  <xs:element name="employee" xdb:defaultTable="EMPLOYEE_XML">
    <xs:complexType>
Next step : create a resource, or just directly insert an XML document into the table.
Example of creating a resource :
declare
  res  boolean;
  doc  xmltype := xmltype(
'<employee>
  <details>
    <emp_id>1</emp_id>
    <emp_name>SMITH</emp_name>
    <emp_age>40</emp_age>
    <emp_dept>10</emp_dept>
  </details>
</employee>'
begin
  res := dbms_xdb.CreateResource(
           abspath   => '/public/test.xml'
         , data      => doc
         , schemaurl => 'my_schema.xsd'
         , elem      => 'employee'
end;
The resource has to be schema-based so that the default storage mechanism is triggered.
It could also be achieved if the document possesses an xsi:noNamespaceSchemaLocation attribute :
SQL> declare
  2 
  3    res  boolean;
  4    doc  xmltype := xmltype(
  5  '<employee xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
  6             xsi:noNamespaceSchemaLocation="my_schema.xsd">
  7    <details>
  8      <emp_id>1</emp_id>
  9      <emp_name>SMITH</emp_name>
10      <emp_age>40</emp_age>
11      <emp_dept>10</emp_dept>
12    </details>
13   </employee>'
14   );
15 
16  begin
17    res := dbms_xdb.CreateResource(
18             abspath   => '/public/test.xml'
19           , data      => doc
20           );
21  end;
22  /
PL/SQL procedure successfully completed
SQL> set long 5000
SQL> select * from "employee1121_TAB";
SYS_NC_ROWINFO$
<employee xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:noNamespaceS
  <details>
    <emp_id>1</emp_id>
    <emp_name>SMITH</emp_name>
    <emp_age>40</emp_age>
    <emp_dept>10</emp_dept>
  </details>
</employee>
Then use XMLTABLE to shred the XML into relational format :
SQL> select x.*
  2  from "employee1121_TAB" t
  3     , xmltable('/employee/details'
  4         passing t.object_value
  5         columns emp_id   integer      path 'emp_id'
  6               , emp_name varchar2(30) path 'emp_name'
  7       ) x
  8  ;
                                 EMP_ID EMP_NAME
                                      1 SMITH