Print Page - how to get a is_DataGrid_Filter_Row true/false property out-of the datagrid?

Using "DataGrid Filter Row" extension, I want to toggle the Filter Row on and off. How can I do that?
My code turning the filter row on and off works and it looks like this:

Code:

function dg_clickHeaderMenu(target) {
	menu.menu('appendItem', { text: 'Filter', handler: function () { dg.datagrid('enableFilter'); } });
	menu.menu('appendItem', { text: 'No filter', handler: function () { dg.datagrid('disableFilter'); } });
	menu.menu('show', {left: 25 + $(target).offset().left, top: -(0) - 10 + $(target).offset().top, align: 'right' });
}

However I want to make it conditionally so either "Filter" or "No filter" is shown dependant on if the filter is there.

So what I would like to have is something similar to:

Code:

function dg_clickHeaderMenu(target) {
	var filter = ???
	if (filter) {
		menu.menu('appendItem', { text: 'No filter', handler: function () { dg.datagrid('disableFilter'); } });
	} else {
		menu.menu('appendItem', { text: 'Filter', handler: function () { dg.datagrid('enableFilter'); } });
	}
}
	
	menu.menu('show', {left: 25 + $(target).offset().left, top: -(0) - 10 + $(target).offset().top, align: 'right'	});
}

where my question is - how do I get the "DataGrid Filter Row" true / false property out-of the datagrid plugin?

EasyUI Forum

General Category => EasyUI for jQuery => Topic started by: WizPS on December 07, 2019, 03:42:47 PM