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Hello.
This is my code for upload:
and I have this grid:
When submitChanges from Grid is called I need to send another page informations. Then I did this:
Now, I need to add my Upload field together.
I've tried:
e.values.UploadField = $('#UploadField').data('tUpload').value();
e.values.UploadField = $('#UploadField').data('tUpload');
But didn't work.
Please, How can I do this?
This is my code for upload:
<%= this.Html.Telerik().Upload() .Name("file") .ShowFileList(false) .Multiple(false) .ClientEvents(events => events.OnSelect("onSelect"))%>and I have this grid:
<%= this.Html.Telerik().Grid<ModelClass>() .Name("Grid")... .ClientEvents(events => { events.OnSubmitChanges("onSubmitChangesGrid"); }) .Editable(editing => editing.Mode(GridEditMode.InCell)) .KeyboardNavigation()%>When submitChanges from Grid is called I need to send another page informations. Then I did this:
function onSubmitChangesGrid(e) { e.values.GeneralName = $('#GeneralName').val(); e.values.Gender = $('#Gender').data('tDropDownList').value(); e.values.Number = $('#Number').data('tTextBox').value(); e.values = $.extend(e.values);}Now, I need to add my Upload field together.
I've tried:
e.values.UploadField = $('#UploadField').data('tUpload').value();
e.values.UploadField = $('#UploadField').data('tUpload');
But didn't work.
Please, How can I do this?