Forum
02:35
04/11/2013
Currently if you initialize the pager as a jQuery object, jqGrid will take the Id attribute of that object and attempt to find the pager by searching the DOM with $("#thatPagerId"), this means you can't build a grid with a pager before inserting it into the DOM, keeping the reference to the pager div as exactly the jQuery object you handed in seems to make a little more sense here. Anyone else agree?
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