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jQuery UI Widgets Forums Grid Grid rendering issue with virtual mode enabled

This topic contains 2 replies, has 2 voices, and was last updated by  maximus2014 11 years, 4 months ago.

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  • maximus2014
    Participant

    Hi,
    I am using jqxgrid-4.5.4. I have requirement for server-side paging, filtering and sorting. I have d3 bar chart. Filtering is triggered when the select a particular portion of the d3 chart. As server-side pagination is a requirement I added ‘virtualmode:true’ option in my grid. While filtering with virtualmode true, filtering is not working properly. All the data from the page load default filtering is displayed. When I make virtualmode false, filtering works perfectly. Could please tell me how to filter correctly with virtualmode true as virtualmode is needed for my server-side pagination.
    Thanks in advance.


    Peter Stoev
    Keymaster

    Hi maximus2014,

    In virtual mode, the filtering should be implemented by the developer, too. The Grid when it is in Virtual Mode, displays only records which you return in the “rendergridrows” callback function. Please ,look at the PHP Integration demos which illustrate how to use the Grid in Server Paging, Filtering & Sorting scenarios.

    Best Regards,
    Peter Stoev

    jQWidgets Team
    http://www.jqwidgets.com


    maximus2014
    Participant

    Thanks Peter for your reply. I’ve implemented filtering with virtualmode true by modifying the “rendergridrows” callback function.

    Old code:

    rendergridrows: function () {
    return dataAdapter.records;
    },

    Corrected one:

    rendergridrows: function (obj) {
    return obj.data;
    },

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