jQuery UI Widgets Forums Grid Data source ID lost when virtual mode on

This topic contains 1 reply, has 2 voices, and was last updated by  Peter Stoev 11 years, 5 months ago.

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  • Data source ID lost when virtual mode on #46581

    dpibt
    Participant

    Hello,

    I have a method to set rows by column id value setted in data source. My source is like that:

    var source = {
        datatype: "json",
        id: columnName,
        datafields: datafields,
        localdata: data
    };

    My function to select rows by source id has one parameter, an array of values that match the value of the column I use as id.

    function setselectedRowsByValue(arrValues){
    	var index = -1;
    	this.value = '';
    	for(var i=0;i<arrValues.length;i++){
    		index = $('#grid').jqxGrid('getrowboundindexbyid', arrValues[i]);
    		this.setselectedRow(index); //own method
    	}
    }

    When virtual mode is off, this method works fine and selects all the rows I pass in the parameter array (in multiplerows selection mode). When I set virtual mode ON, this function stops working and returns -1 always.

    Does anybody know why?

    Thank you!

    Data source ID lost when virtual mode on #46582

    Peter Stoev
    Keymaster

    Hi dpibt,

    It is not possible to select something which is Virtual by value. In Virtual Mode, the Grid displays the data returned by the “rendergridrows” callback function implemented by you. There is no bound IDs and no bound rows in that mode. It is not possible to select items by values which are virtual and are changed all the time.

    Best Regards,
    Peter Stoev

    jQWidgets Team
    http://www.jqwidgets.com

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