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我想用我以前用过一百次的代码制作幻灯片，但由于某种原因它现在不起作用。我无法弄清楚我做错了什么。HTML

<div id="project1" class="project curent">
 <img src="http://images.nationalgeographic.com/wpf/media-live/photos/000/004/cache/african-lion-male_436_600x450.jpg" class="active" />
 <img src="http://www.howdoeslooklike.com/wp-content/uploads/2012/10/Lion-013-2048x2048.jpg" />
 <img src="http://us.123rf.com/400wm/400/400/aleksm/aleksm1209/aleksm120900002/15398724-winged-lion-front-view-drawing.jpg" />
</div>

CSS

.project{position:absolute; top:0; left:0; width:100%;}
.project img{position:absolute; top:0; left:0;  width:200px; height:auto;z-index:8;}
.project img.active{z-index:10;}
.project img.last-active{z-index:9;}

jQuery

function slideSwitch() {
    var $active = $(".project img.active");
    $active.hide();
    if ( $active.length == 0 ) $active = $('.project IMG:last');

    var $next =  $active.next().length ? $active.next()
        : $('.project IMG:first');

    $active.addClass('last-active');

    $next.css({opacity: 0.0})
        .addClass('active')
        .animate({opacity: 1.0}, 1000, function() {
            $active.removeClass('active last-active');
        });
}



$(function() {
    setInterval( "slideSwitch()", 1000 );
});

这是一个正在进行的示例。有什么想法吗？:|

4

1 回答 1

3

slideSwitch()在 setInterval 中未正确调用。这显示在控制台中。另请注意，您的幻灯片代码无法正常工作，因此我添加了一个.show()以使.last-active图像再次出现。

jQuery：

function slideswitch() {
    var $active = $("#project1 img.active");
    $active.hide();
    if ( $active.length == 0 ) $active = $('#project1 IMG:last');

    var $next =  $active.next().length ? $active.next()
        : $('#project1 IMG:first');

    $active.addClass('last-active').show();

    $next.css({opacity: 0.0})
        .addClass('active')
        .animate({opacity: 1.0}, 1000, function() {
            $active.removeClass('active last-active');
        });
};
$(document).ready(function() {
    setInterval(slideswitch, 1000 );
});

小提琴：http: //jsfiddle.net/sdGhT/8/

于 2013-02-21T13:06:21.420 回答