Question

我写了以下内容让我从数组中加载脚本：

scriptLoader = (function(){

  var loader = function (src, handler) {
    if(!self.errorstate){
        var script = document.createElement("script");
        script.src = src;
        script.onload = script.onreadystatechange = function () {
            script.onerror = loaderror(src);
            script.onreadystatechange = script.onload = null;
        };
        var head = document.getElementsByTagName("head")[0];
        (head || document.body).appendChild(script);
    }
  };

  var loaderror = function (src) {
    self.errorstate = true;
    console.log('Error: '+src);
    return false;
  };

  var self = {};
  self.errorstate = false;
  self.files = null;
  self.callback = null;
  self.go = function(){
    if (self.files.length !== 0) {
        loader(self.files.shift(), arguments.callee);
    } else {
        if (self.callback && typeof (self.callback) === 'function' && !self.errorstate) {
            self.callback();
        }
    }
  };
  return self;

});

我正在打电话：

var scripts = [
    'js/placeholder.js',
    'js/modal.js',
    'js/alert.js',
    'js/notafile.js'
];


loader = new scriptLoader();
loader.files = scripts;
loader.callback = alert('loaded');
loader.go();

一切似乎都有效，除了我无法让它在出错（js/notafile.js）时中断 - 所以停止加载任何更多的脚本并且不允许回调触发。它在控制台中显示错误信息，我只是找不到阻止它的好方法！

Answer 1

使用try / catch + throw

http://www.w3schools.com/js/js_errors.asp

var loaderror = function (src) {
    errorstate = true;
    throw 'Error: '+src;
    return false;
};

和

var scripts = [
    'js/placeholder.js',
    'js/modal.js',
    'js/alert.js',
    'js/notafile.js'
];


loader = new scriptLoader();
loader.files = scripts;
loader.callback = alert('loaded');

try{
   loader.go();
}catch(err){
   console.log(err);
}

加载脚本并打破错误

1 个答案: