我试图在页面上创建滑块的多个实例。每个滑块都应该知道它当前正在查看的是哪个幻灯片。似乎当我更新slide
属性时,我为类而不是实例更改了它。这表明我没有在我的公共init()
函数中正确实例化。我哪里做错了?
var MySlider = (function() {
'use strict';
var animating = 0,
slides = 0, // total slides
slider = null,
slide = 0, // current slide
left = 0;
function slideNext(e) {
if ((slide === slides - 1) || animating) return;
var slider = e.target.parentNode.children[0],
x = parseFloat(slider.style.left);
animate(slider, "left", "px", x, x - 960, 800);
slide++;
}
return {
init: function() {
var sliders = document.querySelectorAll('.my-slider'),
l = sliders.length;
for (var i = 0; i < l; i++) {
sliders[i] = MySlider; // I don't think this is correct.
slider = sliders[i];
buildSlider(slider);
}
}
}
})();
根据你的评论,我认为这更像是你想要的:
MySlider = (function () {
Slider = function (e) {
this.e = e;
// other per element/per slider specific stuff
}
...
var sliders; // define this out here so we know its local to the module not init
return {
init: function () {
sliders = document.querySelectorAll('.my-slider');
var l = sliders.length;
for (var i = 0; i < l; i++) {
sliders[i] = new Slider(sliders[i]); //except I'd use a different array
slider = sliders[i];
buildSlider(slider);
}
}
})();
通过这种方式,您将每个元素与它自己的特定元素数据关联起来,但您有一个包含模块,您可以在其中操作您的模块集合
似乎当我更新幻灯片属性时,我为类而不是实例更改了它。
你是正确的。该代码仅在定义MySlider类时运行。如果你想要一个实例变量,你需要在返回的对象中声明它,也就是返回块的一部分:
var MySlider = (function(param) {
return {
slider: param,
init: function() {
var sliders = document.querySelectorAll('.my-slider'),
l = sliders.length;
for (var i = 0; i < l; i++) {
sliders[i] = MySlider; // I don't think this is correct.
slider = sliders[i];
buildSlider(slider);
}
}
});