Use beforeInitData  event and retun false in order not to show the form.

Your code can be optimized if you just call getRowData and use the needed fields insted of calling getCell 3 times.

See docs.

Regards

I have:-

beforeShowForm:

function (form) {

$(

"#isAB").attr('checked', true);

var ajaxData = {};

var list = $("#list1");

var selectedRow = list.jqGrid('getGridParam', 'selrow');

var myCellData = list.jqGrid('getCell', selectedRow, 'AppBudgetStatusID');

var myCellData2 = list.jqGrid('getCell', selectedRow, 'EntityID');

var myCellData3 = list.jqGrid('getCell', selectedRow, 'DeptID');

var ids = list.jqGrid('getGridParam', 'selarrrow');

if(selectedRow==null)

{

//How to to AVOID the updateDialog form being displayed?

}

Please advise.

Thank you.

Regards,

Micheale