Question

I am using a switch-statement to switch between modes on a the jqPlot jQuery plugin.

How can I apply my value from the switch to the function parameters?

I have

type = detectType(gType);

I need this to display, where type is in the parameters below. It does not return that value:

function detectType(value){
    switch(value){
        case 'bar':
            return 'renderer:$.jqplot.BarRenderer,';
            break;
        case 'pie':
            return 'renderer: $.jqplot.PieRenderer,';
            break;
    }
}

function generateGraph(id,gType) {
    type = detectType(gType);
    var s1 = [2, 6, 7, 10];
    var s2 = [7, 5, 3, 4];
    var s3 = [14, 9, 3, 8];

$.jqplot(''+id+'', [s1, s2, s3], {
        stackSeries: true,
        captureRightClick: true,
        seriesDefaults: {
            type
            rendererOptions: {
                barMargin: 30,
                highlightMouseDown: true   
            },
            pointLabels: {
                show: true
            }
        },
        axes: {
            xaxis: {
                renderer: $.jqplot.CategoryAxisRenderer
            },
            yaxis: {
                padMin: 0
            }
        },
        legend: {
           show: true,
           location: 's',
           placement: 'outside'
        }
    });
}

Answer 1

You can’t do what you’re trying to do with the type string. It’s invalid syntax.

Instead, detectType should just return the renderer object itself, not as a string. Then, assign that to renderer in your series defaults. Like so:

function detectType(value){
    switch(value){
        case 'bar':
            return $.jqplot.BarRenderer;

        case 'pie':
            return $.jqplot.PieRenderer;
    }
}

Then:

//...
seriesDefaults: {
    renderer: type
    //...
}
//...