Generating structured XML file based on relational data base

Question

Hi,
I need to use XML DB to generate a structured XML file based on the relational oracle database. I start by using Oracle default table, emp,
Here is EMP data,
EMPNO ENAME DEPNO SAL
123.00 E1 20.00 1,000.00
124.00 E2 20.00 2,000.00
125.00 E3 20.00 2,000.00
126.00 E4 30.00 3,000.00
127.00 E5 30.00 3,000.00
128.00 E6 30.00 4,000.00
129.00 E7 40.00 7,000.00
I used this SQL statement to generate an XML output,
select XMLElement ("Department",
XMLAttributes(deptno as "DEPARTMENTNO"),
XMLElement("EmployeeName", ename),
XMLElement("Salary", sal)
) "Result"
from
emp
where
deptno=20
The result was
<Department DEPARTMENTNO="20">
<EmployeeName>E1</EmployeeName>
<Salary>1000</Salary>
</Department>
<Department DEPARTMENTNO="20">
<EmployeeName>E2</EmployeeName>
<Salary>2000</Salary>
</Department>
<Department DEPARTMENTNO="20">
<EmployeeName>E3</EmployeeName>
<Salary>2000</Salary>
</Department>
but I need this structure instead,
<Department DEPARTMENTNO="20">
<EmployeeName>E1</EmployeeName>
<Salary>1000</Salary>
<EmployeeName>E2</EmployeeName>
<Salary>2000</Salary>
<EmployeeName>E3</EmployeeName>
<Salary>2000</Salary>
</Department>
Could you guide me how I can generate this kind of structure like master-detail structure and which document you recommend to obtain this point.
Thanks,
Shiva

Answer 1

Hi,
If you want to write an xml into file, I think you should use PL/SQL.
Is it even possible to spool output into file without the query itself?
doen't show me a proper xml fileTo get well-formed xml, you also have to make a root element.
If you want <?xml version="1.0"?> also then use xmlroot(), but I haven't figure out
how to specify the encoding.
SQL> WITH xtab AS(SELECT 'E1' ename,20 depno,1000 sal FROM dual
  2               UNION ALL
  3               SELECT 'E2',20,2000 FROM dual
  4               UNION ALL
  5               SELECT 'E3',40,3000 FROM dual
  6               UNION ALL
  7               SELECT 'E4',30,4000 FROM dual)
  8  SELECT XMLRoot(XMLElement("Company",XMLAgg(XMLElement("Department",
  9           XMLattributes(depno as "DEPARTMENTNO"),
10             XMLAgg(XMLForest(
11                      ename AS "EmployeeName",
12                      sal   AS "Salary"))))),version '1.0') company_xml
13  FROM xtab
14  GROUP BY depno;
COMPANY_XML                                                                    
<?xml version="1.0"?>                                                          
<Company>                                                                      
  <Department DEPARTMENTNO="20">                                               
    <EmployeeName>E1</EmployeeName>                                            
    <Salary>1000</Salary>                                                      
    <EmployeeName>E2</EmployeeName>                                            
    <Salary>2000</Salary>                                                      
  </Department>                                                                
  <Department DEPARTMENTNO="30">                                               
    <EmployeeName>E4</EmployeeName>                                            
    <Salary>4000</Salary>                                                      
  </Department>                                                                
  <Department DEPARTMENTNO="40">                                               
    <EmployeeName>E3</EmployeeName>                                            
    <Salary>3000</Salary>                                                      
  </Department>                                                                
</Company>
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