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jQuery UI Widgets Forums Grid jqxPivotGrid setColumnWidth creates error (bug?)

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This topic contains 2 replies, has 2 voices, and was last updated by  ewoudwijma 7 years, 6 months ago.

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  • Author

  • ewoudwijma
    Participant

    After adding setColumnWidth to each row of the pivottable I get an error about a variable called desi not known.

    The code:
    var pivotGridInstance = $(‘#divPivotGrid’).jqxPivotGrid(‘getInstance’);
    var pivotRows = pivotGridInstance.getPivotRows();
    for (i=0; i<pivotRows.items.length; i++)
    {
    pivotRows.items[i].expand();
    pivotRows.setColumnWidth(i, 200);
    }
    pivotGridInstance.refresh();

    After debugging jqxPivotGrid.js I found that a variable is used but nowhere declared:
    b=c>e?desi:e

    I replaced it with b=c>e?c:e which made the error disappear.

    @JQWidget team, can you please check if this is indeed the solution and if so add this in future version.

    Thx,
    Ewoud


    Peter Stoev
    Keymaster

    We are not aware how to reproduce this. Share a jsfiddle.net sample please, which demonstrates an issue with our current version. You are sharing minified code, too.

    Best Regards,
    Peter Stoev

    jQWidgets Team
    https://www.jqwidgets.com/


    ewoudwijma
    Participant

    Hi Peter, Thx for your comment. I will create a jsfiddle.net to show it. Is there an guideline how to do this for JQWidgets? I will do it in about 4 weeks as I will be offline until then.

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