JSONP with knockout

[lev_povolotskyParticipant]

Hi,

Can anyone help me to create a grid that take the data from other source(JSONP) and to use knockout for retrieving because I’m going to retrieve more then 15000, and I don’t want the the wait logo will be for a period of time.

I want that I will get the first data imminently and then it will continue to enrich the grid.

Thanks in advance.

[Peter StoevKeymaster]

Hi lev_povolotsky,

If you want to load such amounts of data, I suggest you to implement Server Paging i.e to enable the Grid’s virtualmode and to implement the rendergridrows callback function. Example: http://www.jqwidgets.com/jquery-widgets-demo/demos/jqxgrid/virtualdata.htm?arctic and http://www.jqwidgets.com/jquery-widgets-demo/demos/php/serverpaging.htm?arctic. With Knockout, make sure to use the approach with jqxDataAdapter(http://www.jqwidgets.com/jquery-widgets-demo/demos/jqxknockout/gridwithjson.htm?arctic).

Hope this helps.

Best Regards,
Peter Stoev

jQWidgets Team
http://www.jqwidgets.com/

[lev_povolotskyParticipant]

According to 3rd link

I don’t succeed to parse jsonp result.
$.parseJSON isn’t working

as you can see here in your esample

 var url = "../sampledata/beverages.txt";

            var GridModel = function () {
                this.items = ko.observableArray();
                var me = this;
                $.ajax({
                    datatype: 'json',
                    url: "../sampledata/beverages.txt"
                }).done(function (data) {
                    var jsonData = $.parseJSON(data);
                    me.items(jsonData);
                });
            };

[Peter StoevKeymaster]

$.parseJSON is a jQuery function and according to me it works if you try to parse Valid JSON String.

Best Regards,
Peter Stoev

jQWidgets Team
http://www.jqwidgets.com/

[lev_povolotskyParticipant]

but I try to parse JSONP, and founded that I shouldn’t parse it just need to insert it.

moreover I succeed to load the data into me.items(data) and it’s works.

The issue now that it doesn’t repreasnt me the data and I don’t know why

I use exactly these code but diffrent columns and it doesn’t represnt me it and I don’t have any error
`
var model = new GridModel();
// prepare the data
var source =
{
datatype: “observablearray”,
datafields: [
{ name: ‘name’ },
{ name: ‘type’ },
{ name: ‘calories’, type: ‘int’ },
{ name: ‘totalfat’ },
{ name: ‘protein’ },
],
id: ‘id’,
localdata: model.items
};
var dataAdapter = new $.jqx.dataAdapter(source);
$(“#grid”).jqxGrid(
{
width: 670,
source: dataAdapter,
columns: [
{ text: ‘Name’, datafield: ‘name’, width: 250 },
{ text: ‘Beverage Type’, datafield: ‘type’, width: 250 },
{ text: ‘Calories’, datafield: ‘calories’, width: 180 },
{ text: ‘Total Fat’, datafield: ‘totalfat’, width: 120 },
{ text: ‘Protein’, datafield: ‘protein’, minwidth: 120 }
]
});
ko.applyBindings(model);
`

Do you know why it’s happened?

[Author]

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