我在这里创建了一个解决方案:
var sliderObjects = [];createSliderObjects();function plusDivs(obj, n) { var parentDiv = $(obj).parent(); var matchedDiv; $.each(sliderObjects, function(i, item) { if ($(parentDiv[0]).attr('id') == $(item).attr('id')) { matchedDiv = item; return false; } }); matchedDiv.slideIndex=matchedDiv.slideIndex+n; showDivs(matchedDiv, matchedDiv.slideIndex);}function createSliderObjects() { var sliderDivs = $('.slider'); $.each(sliderDivs, function(i, item) { var obj = {}; obj.id = $(item).attr('id'); obj.divContent = item; obj.slideIndex = 1; obj.slideContents = $(item).find('.mySlides'); showDivs(obj, 1); sliderObjects.push(obj); });}function showDivs(divObject, n) { debugger; var i; if (n > divObject.slideContents.length) { divObject.slideIndex = 1 } if (n < 1) { divObject.slideIndex = divObject.slideContents.length } for (i = 0; i < divObject.slideContents.length; i++) { divObject.slideContents[i].style.display = "none"; } divObject.slideContents[divObject.slideIndex - 1].style.display = "block";}<link href="http://www.w3schools.com/lib/w3.css" rel="stylesheet"/><script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script><h2 >Manual Slideshow</h2><div id="div1"> <img src="https://i.imgur.com/eLarayS.jpg" > <img src="https://i.imgur.com/xpOiMWh.jpg" > <img src="https://i.imgur.com/lgcC8Y5.jpg" > <img src="http://i.imgur.com/ufmiVTQ.jpg" > <a onclick="plusDivs(this,-1)">❮</a> <a onclick="plusDivs(this,1)">❯</a></div><div id="div2"> <img src="https://i.imgur.com/eLarayS.jpg" > <img src="https://i.imgur.com/xpOiMWh.jpg" > <img src="https://i.imgur.com/lgcC8Y5.jpg" > <img src="http://i.imgur.com/ufmiVTQ.jpg" > <a onclick="plusDivs(this, -1)">❮</a> <a onclick="plusDivs(this, 1)">❯</a></div>您现在可以为带有“ slider”类和唯一ID的滑块添加尽可能多的div。



