jQuery Grid Plugin - jqGridForum

Topic RSS

Promlem with multiply grids with same #id or without #id at all

25/11/2011
19:11

millfreedom

Ukraine

New Member

Members

Forum Posts: 1

Member Since:
25/11/2011

Offline

e.g. I want to create 2 grids:

</div>

</div>

then I try to create it:

jQuery('#grid_one_area #jqGridTable').jqGrid({...some options...});

jQuery('#grid_two_area #jqGridTable').jqGrid({...some options...});

But, HEY!

Now second grid supersedes first!

OK, I try to change selectors:

</div>

</div>

Now, even first grid creation fails:

jQuery('#grid_one_area .jqGridTable').jqGrid({...some options...});

I think that is because jqGrid works incorrect with object's selector it was created from and try to figure out "id" of this object.

Plz, fix it or give advice to me, how can i use two or more grids WITH SAME identifiers (this is mandatory rule).

25/11/2011
19:27

OlegK

Germany

Member

Members

Forum Posts: 1255

Member Since:
10/08/2009

Offline

Hello millfreedom,

You are definitifely wrong in the first example because you used the same id="jqGridTable" twice on one HTML page. The ids must be unique corresponds to HTML/XHTML specification.

The second example, with the usage of <table class="jqGridTable"></table>, is correct as HTML code, but it can't be used by jqGrid. The problem is that jqGrid require id attribute on the table element. The same requirements exists in some jQuery UI Widgets (the datepicker for example). You don't need to use the id in the selector which create jqGrid, but you have to include id attribute to the table elements. So you have to modify you code to the following

Then you can create jqGrid with the code like

jQuery('#grid_one_area .jqGridTable').jqGrid({…some options…});

jQuery('#t_one').jqGrid({…some options…});

Regards
Oleg

All RSS

Forum Timezone: Europe/Sofia

Most Users Ever Online: 715

Currently Online:
41 Guest(s)

Currently Browsing this Page:
1 Guest(s)