document.createElement（'script'）...添加两个脚本与一个回调 [英] document.createElement('script')... adding two scripts with one callback

问题描述

我需要添加原型，然后添加scriptaculous并在加载时得到回调。我目前正在加载原型：

I need to add prototype and then add scriptaculous and get a callback when they are both done loading. I am currently loading prototype like so:

var script = document.createElement("script");
script.src = "http://ajax.googleapis.com/ajax/libs/prototype/1.6.1.0/prototype.js";
script.onload = script.onreadystatechange = callback;
document.body.appendChild( script );

我可以通过链接回调来做到这一点，但这似乎是糟糕的练习（我不想一个愚蠢的20个回调方法链，当我需要加载更多的脚本）。想法？

I could do this by chaining the callbacks, but that seems like poor practice ( I don't want a silly chain of 20 callback methods when I need to load more scripts). Ideas?

推荐答案

我建议您使用一些小型装载机，为您链接并做些事情。例如像这样一个：

I propose you to use some small loader which will chain and do stuff for you. For example like this one:

function loadScripts(array,callback){
    var loader = function(src,handler){
        var script = document.createElement("script");
        script.src = src;
        script.onload = script.onreadystatechange = function(){
            script.onreadystatechange = script.onload = null;
            handler();
        }
        var head = document.getElementsByTagName("head")[0];
        (head || document.body).appendChild( script );
    };
    (function run(){
        if(array.length!=0){
            loader(array.shift(), run);
        }else{
            callback && callback();
        }
    })();
}

此脚本可帮助您构建脚本标记，并在所有文件被加载。调用很简单：

This script should help you to build the script tags and call your callback when all files are loaded. Invoke is pretty easy:

loadScripts([
   "http://ajax.googleapis.com/ajax/libs/jquery/1.3.2/jquery.min.js",
   "http://ajax.googleapis.com/ajax/libs/prototype/1.6.1.0/prototype.js"
],function(){
    alert('All things are loaded');
});

希望这将有助于

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