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

php wont update my db
hello,

sorry for posting in mysql forum but i dont know where exactly is the problem but here

BB_Code error
I'm having a problem with a custom built function and keep getting this error:


Warning: M

Why does my crawler script suddenly end with no error?
Hi.

I have written a web crawler script. It will visit a large number of URL's with cURL.

Calender Not Opening
The following code is not loading the javascript calender in another window. It's not doing anything

preg_match logical error
Code: <?php
$s = file_get_contents("page.html");
preg_match('/<div cla

Chat Box in PHP
I was thinking in doing a Chat Box in PHP. For that I would use a form with two fields, Nick and Mes

Table control is disappearing from Screen
Hi Experts,

I have a screen with multiple table control(TC). Each table control has two c

Help with PHP Calendar code...
Hello, I'm new to this forum and I'm glad I found it.
I wrote this code for a PHP calendar as an

Help on code output
My CODE:

Code: [Select] echo "<phone>".$line["phone"].&qu

Converting RGB values to HEX
Code: <?PHP

$file_handle = fopen("colors/rgb.csv", "r");

wh

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