country state city dropdown in php

Question

Hi! I am new to php and I want to know about country state city drop down populate from database.I searched it for everywhere most of them are used with javascript and some of them are don't work.please help me.
Can someone tell me about this in php
http://www.c-sharpcorner.com/uploadfile/satyapriyanayak/dfgd/
I'll appreciate it.
Thanks for your time!

Satyapriya Nayak · Answer

Hi Wailin Table Creation -- phpMyAdmin SQL Dump -- version 2.10.1 -- http://www.phpmyadmin.net -- -- Host: localhost -- Generation Time: Feb 17 2013 at 09:49 AM -- Server version: 5.0.45 -- PHP Version: 5.2.5 SET SQL_MODE=NO_AUTO_VALUE_ON_ZERO; -- -- Database: `Work` -- -- -------------------------------------------------------- -- -- Table structure for table `city` -- CREATE TABLE `city` ( `CityId` varchar(255) NOT NULL `State_Id` int(11) NOT NULL `CityName` varchar(255) NOT NULL ) ENGINE=MyISAM DEFAULT CHARSET=latin1; -- -- Dumping data for table `city` -- INSERT INTO `city` (`CityId` `State_Id` `CityName`) VALUES ('1' 1 'Bhubaneswar') ('2' 2 'Hyderbad') ('3' 3 'Kolkata') ('4' 4 'Mumbai') ('5' 5 'Calgary') ('6' 5 'Edmonton') ('7' 5 'Duffield') ('8' 6 'Phoenix') ('9' 7 'San Diego') ('10' 8 'Austin') ('11' 9 'Toronto'); -- -------------------------------------------------------- -- -- Table structure for table `country` -- CREATE TABLE `country` ( `Country_Id` varchar(255) NOT NULL `CountryName` varchar(255) NOT NULL ) ENGINE=MyISAM DEFAULT CHARSET=latin1; -- -- Dumping data for table `country` -- INSERT INTO `country` (`Country_Id` `CountryName`) VALUES ('1' 'India') ('2' 'Usa') ('3' 'Canada'); -- -------------------------------------------------------- -- -- Table structure for table `state` -- CREATE TABLE `state` ( `StateId` varchar(255) NOT NULL `Country_Id` int(11) NOT NULL `StateName` varchar(255) NOT NULL ) ENGINE=MyISAM DEFAULT CHARSET=latin1; -- -- Dumping data for table `state` -- INSERT INTO `state` (`StateId` `Country_Id` `StateName`) VALUES ('7' 2 'California') ('1' 1 'Orissa') ('2' 1 'Andhra Pradesh') ('3' 1 'West Bengal') ('4' 1 'Maharastra') ('5' 3 'Alberta') ('6' 2 'Arizona') ('8' 2 'Texas') ('9' 3 'Ontario'); Now let's move to Coding part config.php $host=localhost; $username=root; $password=; $dbname=Work; $con=mysql_connect($host$username$password); mysql_select_db($dbname$con); ?> Country State City.php include('config.php'); ?> Untitled Document function showState(Country_Id) { document.frm.submit(); } function showCity(State_Id) { document.frm.submit(); } Country --Select-- $sql1=select * from country; $sql_row1=mysql_query($sql1); while($sql_res1=mysql_fetch_assoc($sql_row1)) { ?>  > } ?> State --Select-- $sql=select * from state where Country_Id='$_REQUEST[Country_Id]'; $sql_row=mysql_query($sql); while($sql_res=mysql_fetch_assoc($sql_row)) { ?>  > } ?> City --Select-- $sql=select * from city where State_Id='$_REQUEST[state_id]'; $sql_row=mysql_query($sql); while($sql_res=mysql_fetch_assoc($sql_row)) { ?> > } ?> Thanks If this post helps you mark it as answer

Sona Kundnani · Answer

This code will store n keep the value in your combobox.. $(#state_id).val( );

wailin tun · Answer

Thanks a lot bro! your code is work but there's something wrong when I select one of state from state drop down it change back to ---select---. what should I do?please help me again. thanks for your time!