Question

我找到了非常有趣的jQuery Cycle插件。

因此，有以下脚本应该返回图像：

$(function() { 
// retrieve list of slides from server 
$.getJSON('slidelist.php', startSlideshow); 

function startSlideshow(slides) { 
    /* server returns an array of slides which looks like this: 
    [ 
        'images/beach2.jpg', 
        'images/beach3.jpg', 
        'images/beach4.jpg', 
        'images/beach5.jpg', 
        'images/beach6.jpg', 
        'images/beach7.jpg', 
        'images/beach8.jpg' 
    ] 
    */ 

    var totalSlideCount = 1 + slides.length; 

    var $slideshow = $('#slideshow'); 

    // markup contains only a single slide; before starting the slideshow we  
    // append one slide and prepend one slide (to account for prev/next behavior) 
    $slideshow.prepend('<img src="'+slides.pop()+'" />'); 
    $slideshow.append('<img src="'+slides.shift()+'" />'); 

    // start slideshow 
    $('#slideshow').cycle({ 
        fx: 'scrollHorz', 
        startingSlide: 1,  // start on the slide that was in the markup 
        timeout:  0, 
        speed:    500, 
        prev:    '#prev', 
        next:    '#next', 
        before:   onBefore 
    }); 


    function onBefore(curr, next, opts, fwd) { 
        // on Before arguments: 
        //  curr == DOM element for the slide that is currently being displayed 
        //  next == DOM element for the slide that is about to be displayed 
        //  opts == slideshow options 
        //  fwd  == true if cycling forward, false if cycling backward 

        // on the first pass, addSlide is undefined (plugin hasn't yet created the fn yet) 
        if (!opts.addSlide) 
            return; 

        // have we added all our slides? 
        if (opts.slideCount == totalSlideCount) 
            return; 

        // shift or pop from our slide array  
        var nextSlideSrc = fwd ? slides.shift() : slides.pop(); 

        // add our next slide 
        opts.addSlide('<img src="'+nextSlideSrc+'" />', fwd == false); 
    }; 
};

}）;

但是，因为我在ajax中非常糟糕，你能告诉我如何使用这个

$.getJSON('slidelist.php', startSlideshow);

返回图像我该如何收到这些图片网址？

非常感谢，问候。

Answer 1

嗯..根本不是关于AJAX：）

问题是你在函数startSlideshow中收到了一个对象：

slides = [ 
    'images/beach2.jpg', 
    'images/beach3.jpg', 
    'images/beach4.jpg', 
    'images/beach5.jpg', 
    'images/beach6.jpg', 
    'images/beach7.jpg', 
    'images/beach8.jpg' 
];

现在，您可以使用基本的for循环迭代此对象：

for(var i=0, len = slides.length; i<len;i++){
        console.log(slides[i])
}

现在，每一步都有一个图片网址：slides[i]。您可以附加到$('#slideshow')并开始玩游戏。

使用JQuery的JQuery Cycle和JSON

1 个答案: