Can't seem to capture a variable in a chained select


Posted on 16th Feb 2014 07:03 pm by admin

I'm *this* close to having a chained select running but for some reason it doesn't seem to be picking up a variable.

Code: <?php

Did you know?Explore Trending and Topic pages for more stories like this.
require ('inc/connection.php');

//seeming that we are just submitting and refreshing to the one page we need to check if the post variable is set, and if so a couple of other variables are set
if(!isset($_POST['state'])) {
$next_dropdown = 0;
}
else {
//When set this variable reveals the next drop down menu
$next_dropdown = 1;
//this variable keeps the previous selection selected
$selected = $_POST['state'];
}
?>

<form name="form" method="post" action="">
<select name="state" style="font-size:20px;">
<option value="NULL">State</option>
<?php
$query = "SELECT id, name FROM state ORDER BY name ASC";
$result = mysql_query($query);
while($row = mysql_fetch_array($result))
{?>
<option value="<?php echo $row[0]; ?>" onClick="document.form.submit()" <?php if(isset($selected) && $row[0] == $selected) {echo "selected='selected'";} ?>><?php echo $row[1]; ?></option>n";
<?php }
echo '</form>n';
//this is where the other form will appear if the previous form is submitted
if($next_dropdown == 1) {?>
<form name="form2" action="" method="post">
<select name="city">
<option value="NULL">City</option>
<?php
$query2 = "SELECT * FROM city WHERE state_id = " . $row[0];
$result2 = mysql_query($query2);
while($row2 = mysql_fetch_array($result2))
{ ?>
<option value="<?php echo $row2[0]; ?>" onClick="document.form2.submit()"><?php echo $row2[1]; ?></option>
<?php }?>
</select>
</form>
<?php }
?>

The state drop down works fine. Once a state is selected, it will display the city drop down. However, the city drop down never populates. It's as though it forgets what $row[0] is. Any thoughts?
No comments posted yet

Your Answer:

Login to answer
83 Like 35 Dislike
Previous forums Next forums
Other forums

Count Session and Trigger Events
I am New in PHP, seeking a method to count logged users by counting the sessions or any …, is

login page does not execute a else statement
I've created a login page using sessions.
When an incorrect user name or password is entered then

Array help
Hello i got this code to fetch data from database but it is not working it displays
7
Array

php + mysql count consecutive data
I have a database of values and I want to work out how to display them if the values match a consecu

An odd assignment statement. Can someone explain this assignment to me?
What purpose is served by the bit of code between the two equal signs in the $installurl set? Is th

PHP Mysql Staff Induction System
Hi there, I'm pretty new to PHP and Mysql so could really do with being pointed in the right directi

Email Processor
I have a few questions so this post will be a larger one! Sorry, but I'm a bit of a PHP newbie so be

Help with forum quoting?
Hi im working on a forum and I have alomost finished it but i want a user quote system like twitter

php libs/ browsercap.in
ok i have a host that refuses to stay current. they control my php settings and libs. is there a way

PHP Code / Script To check weather the given email exists in a domain
Hi,

I want to implement the following in my web page

in sign up we will ask to enter u

Sign up to write
Sign up now if you have flare of writing..
Login   |   Register
Follow Us
Indyaspeak @ Facebook Indyaspeak @ Twitter Indyaspeak @ Pinterest RSS



Play Free Quiz and Win Cash