Is it possible to have a dynamic (not hard coded) search filter values for the jqGrid column?
So in the example such as:
$('#my-grid').jqGrid({
datatype: 'json',
loadonce: true,
jsonReader: common.jqgrid.jsonReader('Workorder'),
mtype: 'POST',
colNames: ['Project', 'PO Number', 'Type', 'Folder'],
colModel: [
{ name: 'Project', index: 'Project', width: 80, sortable: false, search:false},
{ name: 'PONumber', index: 'PONumber', width: 60, sortable: false, search: true },
{ name: 'Type', index: 'Type', width: 60, sortable: false, search: true},
{ name: 'Folder', index: 'Folder', width: 60, sortable: false, search: false },
],
scroll: true,
});
I would like the type to have a drop down filter with values that are array of distinct values from the data subset coming back.
How would I achieve this?
Edit
Is the jqGrid data accessible directly? I am looking for something like
Data.Cols[2].Distinct
that will give me the distinct array of values from column 3(in this case). Is this possible?
Edit 2
This is the code:
onLoadComplete: function (data) {
var $this = $('#gridReport');
if ($this.jqGrid("getGridParam", "datatype") === "json") {
// first loading from the server
// one can construct now DISTINCT for 'Type' column and
// set searchoptions.value
$this.jqGrid("setColProp", "Type", {
stype: "select",
searchoptions: {
value: buildSearchSelect(getUniqueNames("Type")),
sopt: ["eq", "ne"]
}
});
}
},