Thank you for your reply!

I've understood my mistake: it was that I set volatile to true in two places
I didn't have to.

Backing to the normal behavior  now I can also provide the component without
its own form.
Now I have another question: How can I pervent the validation of components
outside the grid when I add a row ?

kiuma

On 1/27/07, RonPiterman <[EMAIL PROTECTED]> wrote:

first, I would not in any cas try to write my own form component.

So, stale links:

In order for tapestry to submit a form, all form components must be
rendered on submit on the same order they were rendered for the response
which generated the form.

to make sure this happens, tapestry records the order of the components,
and if it doesn't match, generates a stale link.

why doesn't it match?

say you have a list of 1 object, and you iterate over it and generate an
input box for each item.

If when submitting, the list has suddenly 2 objects, two input boxes are
rendered, one too much...

to prevent this, certain approaches were taken, including the For bean,
If bean ( i think...) and action listeners instead of "listener"
listeners.

So - when doing thing in the right order and persisting the information
in the right place, stale link exceptions can be avoided - show me yours
and I'll show you mine (code)...

Cheers,
Ron





Andrea Chiumenti wrote:
> Hello,
> I'm using Tapestry 4.1.1 and I'm trying to update a comopnent.
> This component is a grid, that in normal state hasn't any input
component.
> The first time i call addNew, that adds a new row with input elements
all
> goes well.
> The next time i call this function I have a StaleLinkException.
>
> with this message:
> Rewind of form Home/thisForm expected allocated id #9 to be 'nameField',
> but
> was 'addNew' (requested by component Home/tableForm.addNew).
>
> The call is an ajax call that updates the table containing the row.
>
> How can I prevent the stale link exception, how does Tapestry 4.1.1threats
> stale links ?
>
> Thanks in advance,
> kiuma
>


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