Disable Filter and Sort for an Individual Column

For my ID column I show a button to do an action on the row instead of the ID.  I don’t want them to be able to Filter or Sort on this column.  How can I disable the filter and sort for that column?

Here is my helper call:

@(Html.Infragistics()
	.Grid(Model)
	.ID("grdPallets")
	.Width("100%")
	.Height("500px")
	.PrimaryKey("Pallet_ID")
	.AutoGenerateColumns(false)
	.AutoGenerateLayouts(false)
	.Columns(column =>
	{
		column.For(model => model.Pallet_ID).HeaderText("").Template("<input type='button' class='btnPrintPackingSlip' value='Print Packing Slip' data-id='${Pallet_ID}' onclick='printPackingSlip(this)' />");
		column.For(model => model.Pallet_Number).HeaderText("Number");
		column.For(model => model.Pallet_Quantity).HeaderText("Quantity");
		column.For(model => model.Pallet_DateShipped).HeaderText("Date Shipped");
	})
	.Features(features =>
	{
		features.Sorting().Type(OpType.Remote);
		features.Paging().Type(OpType.Remote);
		features.Filtering().Type(OpType.Remote);
	})
	.DataSourceUrl(Url.Action("GetPallets"))
	.Render()
)

Hello Andrew,

Thank you for posting in our community!

 I would suggest you implementing the ColumnSettings for each of the two features:

@(Html.Infragistics()

.Grid(Model).ID(“grid1”)

.Columns(column =>

{

column.For(x => x.ProductId).HeaderText(“ID”);

column.For(x => x.Name).HeaderText(“Name”);

…

})

.Height(“300px”).Width(“800px”).PrimaryKey(“ProductId”)

.Features(feature =>                                               {

feature.Filtering().Type(OpType.Remote).ColumnSettings(setting =>

{

setting.ColumnSetting().ColumnKey(“ProductId”).AllowFiltering(false).FilterCondition(“equals”);

setting.ColumnSetting().ColumnKey(“Name”).AllowFiltering(true).FilterCondition(“endsWith”);

feature.Sorting().Type(OpType.Remote).ColumnSettings(setting =>

{

setting.ColumnSetting().ColumnKey(“Name”).AllowSorting(true);

setting.ColumnSetting().ColumnKey(“Number”).AllowSorting(false);

});

})

.DataSourceUrl(Url.Action(“PagingGetData”)).DataBind().Render())   

If you have any further questions or if you need additional assistance please let me know.

 

 

Thank you Dimka.  That did exactly what I needed it to.

Hello Andrew,

If you have any additional questions feel free to update this thread.