xquery - Passing results to a a variable -

alright let me begin saying new xquery. started learning 2 days ago. task involves using 3 separate xml files contain tables.

the first file named dependent.xml contains these fields: essn, dependent_name, relationship.

the second file named employees.xml contains these fields: fname, lname, ssn, dno

the third file named department.xml contains these fields: dname, dnumber, mgrssn

so relationships between these tables are: essn = ssn, ssn = mgrssn , dno = dnumber.

i need return list of dependents, along corresponding employee , corresponding manager of employees. code looks far:

{     $d in doc("../company/dependent.xml")//dependent,         $e in doc("../company/employee.xml")//employee     $d/essn = $e/ssn     return       <dependent       name="{ $d/dependent_name }"       fname="{ $e/fname }" lname="{ $e/lname }"       /> } 

this gives me list of dependents , corresponding employee, point.

the problem managers employees names in employee table. need match dno of employees have dependents dnumber in department table in order mgrssn of departments , match mgrssn employee ssn manager names.

my idea store dno of employees dependents i'm getting code in variable , use variable managers departments , use so:

{     $d in doc("../company/department.xml")//department,         $e in doc("../company/employee.xml")//employee     $d/mgrssn = $e/ssn , $d/dnumber = $**variable**     return       <manager       fname="{ $e/fname }" lname="{ $e/lname }"       /> } 

problem have no idea how , realize on separate loop. there way can in 1 loop? if there easier way need i'm open suggestions :)