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

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

Problem with an browser game.
Hello,i just joined that great forum and i got php prob,its kinda freaky...anyway i got browser game

getAlexaRank($url) function not working
I have made a function to get alexa rank
the site is here: http://mytestsite.rack111.com/1

Undefined Variable: PHP_SELF, pls help
Hi,

Im a newbie on PHP / MySQL programming and Im running a script to search one field on my

need to apply an if/else statement to Tim Thumb script
Not sure how to work this. I essentially want to call a variety of image sizes based on which style

Production Order Enterprise Service to Manufacturing Exctn Sys (MES) ???
Hello Experts,

In our current landscape SAP ECC 5.0 is integrated to MES system via PI 7.

Custom list order
Hi there,

I have checked this tutorial and it's great till the point where I want to display

need help in update query
i create a form for update. there are 8 columns in my mysql table. on my main page all the data is r

mysql select with $_get ?
Hi, i have this code:
Code: // If char id is 0 and character dont exist do:
if ($_GET["id

Code doesn't print what i wanted it too. Please help!
Hi there,

I wrote this code to mae it so that in the form before it that the user said their

Could Someone Please Debug This?
I was wondering if someone could debug this script for me. I realize it's not the tidest script (and

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