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.

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

Your Answer:

Login to answer

Other forums

SAP Treasury - Commodities
We are running SAP ECC 6.0 with Treasury Activated.
EA-FIN is also activated (SFW5).

storing video files into mysql in php
hi

i have my video files in my folder ,
i have to store the path of the videos into db an

help with image text
Hi ...

see i need the text of the name and number to start in the center and always be in the

limitations on array_unique()
Does any one know what/how imitations are applied to array_unique? I have an issue where no matter h

Receive Rosettanet Message to SAP 4.6c
Dear all ,

My customer will send the PO details by rosettanet message , Is it

Get content from table into a list, without repeating.
Alright this is kinda an odd thing, so I need some help.

I have a table "quote" wit

I apologise! It's a newbie to PHP feedback form problem!
Hi,

I am a total newbie to PHP. I am also a unemployed and need to get a job. So after 100% r

urldecode question
How would I format this line of code properly?

<?php echo urldecode($_GET['Title']); ?

Multiple submit forms inside a mysql array page
Hi all,

Im having trouble to get the beneath script to work the thing i can't seem to get wor

Why is this query failing?
Why is this not working?
$query = "SELECT * FROM `users` WHERE `userid` = " . $USERID