Unable to retrieve id from dropdown | Member | | Join Date: Feb 2008
Posts: 63
| | |
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.
|  | Moderator | | Join Date: Jun 2007 Location: York, England, with wolves.
Posts: 4,938
| | | re: Unable to retrieve id from dropdown
You only need to include the relevant code.
You've supplied too much.
| | Member | | Join Date: Feb 2008
Posts: 63
| | | re: Unable to retrieve id from dropdown
Heres the relevant code i think where the change needs to be done.
This is the code for drop down box. -
<?php
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require "config.php"; // database connection details
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echo "
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function fillQualification(){
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// this function is used to fill the Qualification list on load
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";
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$q1=mysql_query("select * from qualification");
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echo mysql_error();
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while($nt1=mysql_fetch_array($q1)){
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echo "addOption(document.form1.Qualification, '$nt1[QualificationId]', '$nt1[QualificationName]');";
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}// end of while
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?>
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} // end of JS function
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function SelectSpecial(){
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// ON or after selection of qualification this function will work
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removeAllOptions(document.form1.Specialization);
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addOption(document.form1.Specialization, "", "-Select-", "");
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// Collect all element of Specialization for various QualificationId
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<?php
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// let us collect all QualificationId and then collect all Specialization for each QualificationId
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$q2=mysql_query("select distinct(QualificationId) from specialization");
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// In the above query you can collect QualificationId from qualification table also.
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while($nt2=mysql_fetch_array($q2)){
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echo "if(document.form1.Qualification.value == '$nt2[QualificationId]'){";
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$q3=mysql_query("select SpecializationArea from specialization where QualificationId='$nt2[QualificationId]'");
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while($nt3=mysql_fetch_array($q3)){
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echo "addOption(document.form1.Specialization,'$nt3[SpecializationId]', '$nt3[SpecializationArea]');";
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} // end of while loop
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echo "}"; // end of JS if condition
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}
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?>
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}
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This is the insert statement from where the details go in database. -
mysql_query("INSERT INTO $db_table1(UserName,PhoneNo,MobileNo,YearsExp,MonthsExp,FunctionalArea,KeySkills,Qualification,Specialization,Institute,FunctionalRoles,CurrentEmployer,PreviousEmployers,ProfileTitle,ResumeFile) values ('$_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[PreviousEmployers]','$_POST[ProfileTitle]','$_POST[ResumeFile]')")
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or die(mysql_error());
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