处理jQuery（document）.ready中的错误

问题描述：

我正在开发用于Web框架的JS，并经常与其他开发人员（通常容易出错）的jQuery代码混合使用。不幸的是，他们的jQuery（文档）.ready块中的错误阻止了我的执行。采取以下简单示例：

I'm developing JS that is used in a web framework, and is frequently mixed in with other developers' (often error-prone) jQuery code. Unfortunately errors in their jQuery(document).ready blocks prevent mine from executing. Take the following simple sample:

<script type="text/javascript">
    jQuery(document).ready(function() {
        nosuchobject.fakemethod();       //intentionally cause major error
    });
</script>
<script type="text/javascript">
    jQuery(document).ready(function() {
        alert("Hello!");                 //never executed
    });
</script>

不管先前发生了什么，第二个就绪块不应该执行吗？是否有一种安全的方式来运行jQuery（document）.ready即使出现以前的错误也会运行？

Shouldn't the second ready block execute regardless of what happened in the previous? Is there a "safe" way to run jQuery(document).ready that will run even in the case of previous errors?

编辑：我没有控制/可见性容易出错的块，因为它们是由其他作者编写的并且是任意混合的。

I have no control/visibility over the error-prone blocks as they're written by other authors and mixed in arbitrarily.

答

我还没试过这段代码，但它应该工作（至少，这个想法应该是）。 ~~确保在jquery之后包含它，但是在任何潜在的错误脚本之前。~~（没必要，请参阅注释。）

I haven't tried this code, but it should work (at least, the idea should anyway). ~~Make sure you include it AFTER jquery, but BEFORE any potentially buggy scripts.~~ (Not necessary, see comments.)

var oldReady = jQuery.ready;
jQuery.ready = function(){
  try{
    return oldReady.apply(this, arguments);
  }catch(e){
    // handle e ....
  }
};

处理jQuery（document）.ready中的错误

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