使spin.js微调器相对于目标div的位置

Question

currently the jQuery spin.js spinner has options to position it.

I have used the following config before:

        // set up spin.js vars
    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: 'mySpinner', // 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('testdiv');
    var spinner = new Spinner(opts).spin(target);
    $(target).data('mySpinner', spinner);

However the position shows its relative to its parent.

So currently the structure of:

<div id="container">
  <div id="testdiv">
  </div>
<div>

The spinner will get centered in the containing div and not the targetted div. Is there a way around this?

Answer 1

If you're still looking for a solution, I guess you can use position: relative to make it work, like this:

.mySpinner {
    position: relative !important;
}

This way the spinner will be centered in your testdiv . Here's a working JsFiddle .