store value of ddlc into a variable

[GoogleCodeExporter]

** What steps will reproduce the problem?
the ddlc is perfectly working. i want to use the value of the ddlc after i 
select some items in a variable in php. how do i get it?

** What is the expected output? What do you see instead?

** What is your environment?
-- DropDownCheckList version:
-- jQuery version:
-- jQuery UI version:
-- Browser and version:

** Please provide any additional information below/Please attach sample
html

Original issue reported on code.google.com by naisarge...@gmail.com on 28 Feb 2012 at 5:42

Attachments:

• emailsubscribers.php

[GoogleCodeExporter]

The DDCL tries to have no effect on the underlying operation of the SELECT it 
is based on.  In other words, whatever you are doing for form submission 
processing in your php that can handle the SELECT should continue to work after 
you activate DDCL.

Original comment by womohun...@ittrium.com on 8 Mar 2012 at 2:12

• Changed state: Done
• Added labels: Type-Other
• Removed labels: Type-Defect

[GoogleCodeExporter]

i solved the problem myself the same day. thanks for your help.

Original comment by naisarge...@gmail.com on 9 Mar 2012 at 4:35