Question

59

jqGrid colModel dynamic Searchoptions

rated 0 times [ 61] [ 2] / answers: 1 / hits: 7501 / 11 Years ago, tue, february 11, 2014, 12:00:00

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"]



                }

            });            

        }

    },

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malkajillc

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3 Years ago

answered 11 Years ago neildrews · Accepted Answer

This is how I ended up doing it, I ended up using jquery to populate the drop down directly, as jqgrid filters were not available at any point (if they are, I would love to see a documented example):

 onLoadComplete: function (data) {



        if ($('#mygrid').jqGrid(getGridParam, datatype) === json) {



            //get the distinct types from the data set coming back

            var length = data.length;

            var types = [];



            for (var i = 0; i < length; i++) {

                types.push(data[i]['Type']);

            }



            var uniqueTypes = getUnique(types);



            $('#gridId select[name=Type]').empty().html('<option value=>All</option>');



            for (var i = 0; i < uniqueTypes.length; i++) {

                  $('#gridId select[name=Type]').append($('<option value=' + uniqueTypes[i] + '>' + uniqueTypes[i] + '</option>'));

            }

        }