Unable to retrieve id from dropdown

[agarwalsrushti]

Hi,
Ive made dynamic drop down boxes. On selection of value from first drop down box the values in second drop down are displayed. I ve made this two drop down in a separate file. I want to store the id value of the selected name in the drop down into database. I am able to store the value of first drop down but the value of second drop down is always saved as 0 in the database.
I think its just a small error im am making somewhere. I think the change has to be made on line 232 in insert statement of process.php file
I ve given the code of both my files below plz let me know how it can be fixed.

[Markus]

You only need to include the relevant code.
You've supplied too much.

[agarwalsrushti]

Heres the relevant code i think where the change needs to be done.
This is the code for drop down box.

[code=php]
<?php

require "config.php "; // database connection details
echo "

function fillQualificati on(){
// this function is used to fill the Qualification list on load

";

$q1=mysql_query ("select * from qualification") ;
echo mysql_error();
while($nt1=mysq l_fetch_array($ q1)){
echo "addOption(docu ment.form1.Qual ification, '$nt1[QualificationId]', '$nt1[QualificationNa me]');";
}// end of while
?>
} // end of JS function

function SelectSpecial() {
// ON or after selection of qualification this function will work

removeAllOption s(document.form 1.Specializatio n);
addOption(docum ent.form1.Speci alization, "", "-Select-", "");

// Collect all element of Specialization for various QualificationId

<?php
// let us collect all QualificationId and then collect all Specialization for each QualificationId
$q2=mysql_query ("select distinct(Qualif icationId) from specialization" );
// In the above query you can collect QualificationId from qualification table also.
while($nt2=mysq l_fetch_array($ q2)){
echo "if(document.fo rm1.Qualificati on.value == '$nt2[QualificationId]'){";
$q3=mysql_query ("select SpecializationA rea from specialization where QualificationId ='$nt2[QualificationId]'");
while($nt3=mysq l_fetch_array($ q3)){
echo "addOption(docu ment.form1.Spec ialization,'$nt 3[SpecializationI d]', '$nt3[SpecializationA rea]');";


} // end of while loop
echo "}"; // end of JS if condition

}
?>
}
[/code]

This is the insert statement from where the details go in database.
[code=php]
mysql_query("IN SERT INTO $db_table1(User Name,PhoneNo,Mo bileNo,YearsExp ,MonthsExp,Func tionalArea,KeyS kills,Qualifica tion,Specializa tion,Institute, FunctionalRoles ,CurrentEmploye r,PreviousEmplo yers,ProfileTit le,ResumeFile) values ('$_POST[UserName]','$_POST[PhoneNo]','$_POST[MobileNo]','$_POST[YearsExp]','$_POST[MonthsExp]','$_POST[AreaName]','$_POST[KeySkills]','$_POST[Qualification]','$_POST[Specialization]','$_POST[InstituteName]','$_POST[RoleName]','$_POST[CurrentEmployer]','$_POST[PreviousEmploye rs]','$_POST[ProfileTitle]','$_POST[ResumeFile]')")
or die(mysql_error ());
[/code]