How to display value in drop down list after form has been reloaded


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

Please bear with me as I am very new to php and html.

I have a form with several drop down menus, the one at the top shows a list that when an item is selected fills in the form below including the other drop downs. So when an item is selected the form gets info from a mysql database and then fills in the form. At this point the user can further modify the form and re-submit it to update the database. This means that the drop downs in the form not only need to display the relavent options from the database to reflect the existing choice but also when clicked on show a list to allow the user the change the selection from the list.

Did you know?Explore Trending and Topic pages for more stories like this.
This means that the dropdown needs to:
show the words 'select item' when the form is first loaded and the form is empty
show the current selection when the defined by a selection in the top drodown
show all the available items when the list is dropped down
Curently my drop down does only this:
shows the nothing when the form is first loaded and the form is empty
shows the current selection when defined by a selection in the top drodown
show all the available items when the list is dropped down plus the one that was there before it was clicked on, therefore doubing the option up showing the same option twice

The form is here if you would like to see it in action: http://www.spencercarpenter.co.uk/portfolioAppFiles/simpleForm.php

And a snippet of code from one of th drop downs is here:

Code: <select name="img01" onChange="updateImg(this, 'thumbImg01')">
<option value=-"-1" ><?php echo $row_chosen_pItem['img01']; ?></option>
<?php buildFileList5('uploaded');?>
</select>
the code inside the 'buildFileList5' looks like this:

Code: <?php
function buildFileList5($theFolder)
{
// Execute code if the folder can be opened, or fail silently
if ($contents = @ scandir($theFolder))
{
// initialize an array for matching files
$found = array();
// Create an array of file types
$fileTypes = array('jpg','jpeg','gif','png');
// Traverse the folder, and add filename to $found array if type matches
$found = array();
foreach ($contents as $item)
{
$fileInfo = pathinfo($item);
if (array_key_exists('extension', $fileInfo) && in_array($fileInfo['extension'],$fileTypes))
{
$found[] = $item;
}
}
// Check the $found array is not empty
if ($found)
{
// Sort in natural, case-insensitive order, and populate menu
natcasesort($found);
$selectedImage = ""; //default image
foreach ($found as $filename)
{
if(!empty($_POST['img01']) && $_POST['img01'] == $filename)
{
$sel = "SELECTED";
$selectedImage = $filename;
}
echo "<option value='$filename' $sel>$filename</option>n";
}
}
}
}
?>



I hope that is enough info to explain it.

If anyone could tell how to get the dropdown working properly I would be very gratefull.

Thanks a lot
No comments posted yet

Your Answer:

Login to answer
241 Like 49 Dislike
Previous forums Next forums
Other forums

Change snippet to use CURDATE
I am using time() in the snippet below, I would like to use CURDATE() and have the database setup fo

Session is not saving
I am not trying to do anything too fancy, I am just trying to get some $_SESSION data to save and us

mysql UNION | warning mysql_fetch_array !!
Code: <?php
$i = 0;
$query1="SELECT * FROM `products` WHERE `div

SQl num_rows problem
when i try to count rows from an SQL select i get an warning

Code: [Select]$countviews = mysq

Pro-rata Billing in Periodic Billing for Rental
Hi guys,

We have a situation, where a rental contract starts at the mid of the month say

Cannot Display Array from Select Statement + Login question
Hi,
I'm new to PHP but so far so goog. I was assigend a project and I'm very close to completion.

Shuffle between users ??
I have multiple $users in table. I need to send them $message.

I need to send the next messag

Can't seem to capture a variable in a chained select
I'm *this* close to having a chained select running but for some reason it doesn't seem to be pickin

Embed Video Problem
Hey, thanks for looking!

File to be embedded:

Code: <playlist version="1&q

undefined offset help
Hi All,

I kept getting undefined offset PHP notice for a simple for loop. For eg
$va = arr

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