0

我正在制作一个可以淡入淡出新闻的新闻板。当元素淡出时，内容会被删除，当它淡入时，会出现新的内容。这行得通，只是时机不对。应该只在旧内容消失时出现的新内容，在元素消失时出现。有没有办法等到这完成？

$(document).ready(function() {
    var counter = 0;
    var slides = ['<p>Track All of Your enemys Using The <span>Scout Tool!</span></p>',
                    '<p>Download powerups Using The <span>Pirate Tool!</span></p>',
                    '<p>Gather Badges To Attack others in <span>Raids!</span></p>',
                    '<p>Enjoy Orgainzing Your Profile <span>Join Now!</span></p>'];

    $('#slide-left').click(function() {
        $('#slide').fadeTo('fast', 0);
        $('#slide p').remove();

        if (counter > 0){counter --;}

        $('#slide').append(slides[counter]);
        $('#slide').fadeTo('fast', 1);
    });
    $('#slide-right').click(function() {
        $('#slide').fadeTo('fast', 0);
        $('#slide p').remove();

        if (counter < (slides.length-1)){counter ++;}

        $('#slide').append(slides[counter]);
        $('#slide').fadeTo('fast', 1);

    });

});
4

3 回答 3

5

fadeTo接受第三个参数，即回调函数。这在动画完成时调用。所以你总是可以做类似的事情：

$('#slide').fadeTo('fast', 1, function() {

    // This part is executed after fade

});
于 2013-05-27T17:26:17.503 回答
1

你可以做 

$('#slide').fadeTo('fast', 0,function(){
     $('#slide p').remove();
     if (counter > 0){counter --;}
     $('#slide').append(slides[counter]).fadeTo('fast', 1);
});
于 2013-05-27T17:25:50.643 回答
-1 
$(document).ready(function() {
    var counter = 0;
    var slides = ['<p>Track All of Your enemys Using The <span>Scout Tool!</span></p>',
                    '<p>Download powerups Using The <span>Pirate Tool!</span></p>',
                    '<p>Gather Badges To Attack others in <span>Raids!</span></p>',
                    '<p>Enjoy Orgainzing Your Profile <span>Join Now!</span></p>'];

    $('#slide-left').click(function() {
        $('#slide').fadeTo('fast', function() {
            $('#slide p').remove();
            if (counter > 0){
                counter --;
            }else{
                if (counter < (slides.length-1)){
                    counter ++;
                }
            }
            $('#slide').append(slides[counter]);
            $('#slide').fadeTo('fast', 1);
        });
    });
});

我不知道这是否正确，但我的 5 次发送：

希望我有帮助，

于 2013-05-27T17:27:46.100 回答