Sample Code
in my php
<div id="mainDiv" style="display:none" >
<div id="innerDiv" style="display:none" ></div>
<div id="outerDiv" style="display:none" ></div>
</div>
in my function I display the mainDiv using.
Element.show('mainDiv');
Then I do a Ajax.Updater and the first time I try to load the content
into the innerDiv. I get
innerDiv is not defined in FireBug.
I know what the reason for this is; that in my first call I show the
parent Div "mainDiv" but the innerDIV has not registered with the DOM.
My question is how do I show the parentDiv("mainDiv") and the child
div at the same time so when I do my Ajax.Updater the
childDiv("innerDiv") is registered with the DOM and I can place the
XHTML into that div? After this works I flip the content out between
the two div's. That way one div is displaying the content to the user
and the other div is getting new content.
I
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