将spin.js微调器应用于目标而不是其父级

Question

when positioniung the spin.js spinner you have options. These options say that it positions it relevant to the targets parent. I do not want this.

I wish however for the spinner to position itself inside and centre of the target div.

Heres the basic set up:

        var opts = {
        lines: 13, // The number of lines to draw
        length: 5, // The length of each line
        width: 2, // The line thickness
        radius:5, // The radius of the inner circle
        corners: 1, // Corner roundness (0..1)
        rotate: 58, // The rotation offset
        direction: 1, // 1: clockwise, -1: counterclockwise
        color: '#fff', // #rgb or #rrggbb or array of colors
        speed: 0.9, // Rounds per second
        trail: 100, // Afterglow percentage
        shadow: false, // Whether to render a shadow
        hwaccel: false, // Whether to use hardware acceleration
        className: 'spinner', // The CSS class to assign to the spinner
        zIndex: 2e9, // The z-index (defaults to 2000000000)
        top: '50%', // Top position relative to parent
        left: '50%' // Left position relative to parent
    };

    // show a spinner in place of the icon
    var target = document.getElementById('followdiv');
    var spinner = new Spinner(opts).spin(target);
    $(target).data('spinner', spinner);

Thanks, Any ideas?

Answer 1

#followdiv{
    /* ... */
    position: relative;
}

This says All absolutely positioned elements inside me will be positioned relatively to me .

JS Fiddle Demo