[jQuery] Re: generated content(PHP) into DIV?

On Apr 30, 12:23 pm, Tony <tony.mak...@ymail.com> wrote:
> Hello Everyone!
>
> I have 3 DIV's next to each other. I want that the first div is loaded
> over jquery with a php file which outputs an unordered list in html.
>
> When I click on a list item in the first DIV I want that jquery sends
> the parameters from the list to a second php file for making an
> unordered list which is loaded this time into the second DIV.
>
> The same should happen from the second DIV into the third DIV whith a
> third php file.
>
> All php files connect to a mysql database which has three tables
> (Place, House, Level). The tables have three columns(id, name,
> location). location is always linked to the next table's id (ex.
> "sesame street 1" has id 37 then tabel Place would have folowing
> entry: id=1 name=Stockholm location=37 )
>
> The purpose of this script is to navigate though houses in different
> places. The 3 DIV's contain following.
> DIV 1 = Place (ex. Stockholm, Berlin, Barcelona)
> DIV 2 = House (sesame street 1, sesame street 2, sesame street 3)
> DIV 3 = Level (Floor 1, Floor 2, Floor 3)
>
> I really tried all things in jquery from appendTo to $.ajax but I just
> dont get it working! The three divs are easy to create but getting the
> content dynamically into them seems quite impossible. Please help me,
> I tride really long to solve this problem but I just don't seem to get
> it, hope someone can help me :(

What does your code look like?

if you have a var placeContent = 'HTML string from ajax request';
$('#place').append( placeContent )
or
$('#place').html( placeContent )
or
$(placeContent).appendTo('#place')

should all work fine