jQuery UI Widgets Forums Editors DateTimeInput Date not showing after parent DIV is hidden then visible

This topic contains 3 replies, has 3 voices, and was last updated by  Peter Stoev 10 years, 7 months ago.

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  • Author

  • mesken
    Participant

    Hi Team,

    I create a Date Time inside a DIV that is hidden on pageload but that is shown when the user clicks on a button. But when I show the parent DIV, the date does not appear and behaves strangely. I even tried to empty the parent DIV and recreate it when the user clicks on the button, but no change.
    I even tried this :

    $(‘#MyparentDIV’).show(300);
    $(“#MyparentDIV”).empty();
    $(“#MyparentDIV”).append(‘<div id=”myDateDIV”></div>’);
    $(“#myDateDIV”).jqxDateTimeInput({ width: ‘100%’, height: ’25px’, formatString: ‘yyyy-MM-dd HH:mm:ss’ });

    No result.

    Thanks


    Peter Stoev
    Keymaster

    Hi mesken,

    If you show the parent’s DIV without the animation i.e if you initialize jqxDateTimeInput from a visible DIV tag, then you won’t have any problem. The visible DIV tag is required for performing a correct layout. By visible, I mean with something different than display: none. It still can be visibility:hidden;. Another option is to call the widget’s “refresh” method once the “show” animation has finished.

    Best Regards,
    Peter Stoev

    jQWidgets Team
    http://www.jqwidgets.com


    Gopre400
    Participant

    Hi Peter,
    I have tried refreshing the widget, when the “show” animation is complete. But the widget does not display.

    The Div is hidden by using hidden=”hidden”

    Here is what I’m doing…am I missing something?
    $(‘#liAddNote’).click(function () {
    $(‘#divAddNote’).show(“slow”, function () {
    $(“#txtNoteDate”).jqxDateTimeInput(‘refresh’);
    });
    $(‘#divItemsMenu’).hide();
    $(‘#txtItemDescription’).val($(‘body’).data(‘item’));
    });


    Peter Stoev
    Keymaster

    Hi Gopre400,

    Sorry, but the provided code and information are not sufficient for testing your application scenario. Please, prepare a sample in http://jsfiddle.net/ and share it. As I already wrote here, a visible DIV tag is required for performing a layout. If your DIV tag is not visible, the widget would not be created.

    Best Regards,
    Peter Stoev

    jQWidgets Team
    http://www.jqwidgets.com

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