Hi everyone,
Just for the records I found a more simple solution for this without the
javascript stuff:
I also had problems that the pager did not update or jump to the new
row/page when
clicked on a page and all this packed in a hidden div.
your tml:
Thank you so much Josh, your solution works great! It's also easy to
understand.
Thanks again,
Paul
Josh Canfield wrote:
>
> I whipped something up that will allow you to only submit the form
> when the link is clicked.
>
> Add this to your form containing the grid to keep track of the page
I whipped something up that will allow you to only submit the form
when the link is clicked.
Add this to your form containing the grid to keep track of the page
Add this to your page:
@Property
private int _page;
@Component
private Grid _grid;
void onSuccess() {
_grid.setCurrentPa
I've found a solution (that I feel is horribly inefficient, but is the best
I've got):
I simply put onclick="$('bookDetailForm').submit()" on my checkbox, and that
way each item that's checked results in a hit back to the server to update
the bean. That way, the form is already updated when you
Thanks a lot for the help Josh, I'm still having trouble though.
I've spent hours now and I think I know what I need to do, I just don't know
how to do it:
I need to have the page links Submit my form and then change the page.
Looking at the generated html, it's just a simple anchor tag directly
I don't use a grid so I could be way off, but a quick look at the
GridPager component tells me that it generates links. Since it's a
link your form isn't getting submitted it's just getting the next
page. I imagine there must be a way to make the pager submit your
form. I don't see a direct way to
