Mouseover triggered on absolute positioned div

Posted by Tauren on Stack Overflow See other posts from Stack Overflow or by Tauren
Published on 2010-04-20T20:07:16Z Indexed on 2010/04/20 20:13 UTC
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Objective

Have a small magnifying glass icon that appears in the top right corner of a table cell when the table cell is hovered over. Mousing over the magnifying glass icon and clicking it will open a dialog window to show detailed information about the item in that particular table cell. I want to reuse the same icon for hundreds of table cells without recreating it each time.

Partial Solution

Have a single <span> that is absolutely positioned and hidden. When a _previewable table cell is hovered, the <span> is moved to the correct location and shown. This <span> is also moved in the DOM to be a child of the _previewable table cell. This enables a click handler attached to the <span> to find the _previewable parent, and get information from it's jquery data() object that is used to populate the contents of the dialog.

Here is a very simplified version of my HTML:

<body>
    <span id="options">
        <a class="ui-state-default ui-corner-all">
            <span class="ui-icon ui-icon-search"></span>
            Preview
        </a>
    </span>
    <table>
         <tr>
             <td class="_previewable">
                 <img scr="user_1.png"/>
                 <span>Bob Smith</span>
             </td>
        </tr>
   </table>
</body>

And this CSS:

#options {
    position: absolute;
    display: none;
}

With this jQuery code:

var $options = $('#options');
$options.click(function() {
    $item = $(this).parents("._previewable");
    // Show popup based on data in $item.data("id");
    Layout.renderPopup($item.data("id"),$item.data("popup"));               
});

$('._previewable').live('mouseover mouseout',function(event) {
    if (event.type == 'mouseover') {
        var $target = $(this);
        var $parent = $target.offsetParent()[0];

        var left = $parent.scrollLeft + $target.position().left 
            + $target.outerWidth() - $options.outerWidth() + 1;
        var top = $parent.scrollTop + $target.position().top + 2;

        $options.appendTo($target);
        $options.css({
            "left": left + "px",
            "top": top + "px"
        }).show();
    }
    else {
        // On mouseout, $options continues to be a child of $(this)
        $options.hide();
    }
});     

Problem

This solution works perfectly until the contents of my table are reloaded or changed via AJAX. Because the <span> was moved from the <body> to be a child of the cell, it gets thrown out and replaced during the AJAX call. So my first thought is to move the <span> back to the body on mouseout of the table cell, like this:

    else {
        // On mouseout, $options is moved back to be a child of body
        $options.appendTo("body");
        $options.hide();
    }

However, with this, the <span> disappears as soon as it is mouseover. The mouseout event seems to be called on _previewable when the mouse moves into the <span>, even though the <span> is a child of _previewable and fully displayed within the boundaries of the _previewable table cell. At this point, I've only tested this in Chrome.

Questions

  1. Why would mouseout be called on _previewable, when the mouse is still within the boundaries of _previewable? Is it because the <span> is absolutely positioned?

  2. How can I make this work, without recreating the <span> and it's click handler on each AJAX table referesh?

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