鴨子打jQuery .on來包裝回調在try/catch

Question

我在JavaScript中記錄客戶端異常。爲了獲得堆棧跟蹤，我需要在try/catch中包裝代碼。

我試圖'鴨拳'jQuery的.on()方法，而不是每個回調單獨包裝。

它的大部分工作，但由於某種原因，它是破壞我的jQueryUI排序。如果我註釋掉下面的代碼，排序工作正常。我如何正確攔截on事件，並確保我得到的論據和背景正確？

if(window.jQuery && jQuery.fn.on){ 
    //keep a reference to the original `on` 
    var _on = jQuery.fn.on; 

    //replace `on` with my version 
    jQuery.fn.on = function() { 

    //start from the end, looking for the callback 
    for(var i=arguments.length-1; i>=0; i--){ 

     //is this the function? 
     if(typeof arguments[i] == 'function'){ 

     //keep a reference to the callback 
     var func = arguments[i]; 

     //replace the callback 
     arguments[i] = function(){ 
      try{ 
      //call the original callback 
      func.apply(this, arguments); 
      }catch(e){ 
      ... 
      } 
     }; 
     break; 
     } 
    } 

    //call the original `on` with our new arguments 
    return _on.apply(this, arguments); 
    }; 
} 

Answer 1

嘗試

if(window.jQuery && jQuery.fn.on){ 
    //keep a reference to the original `on` 
    var _on = jQuery.fn.on; 

    //replace `on` with my version 
    jQuery.fn.on = function() { 
     var self = this; 
     //start from the end, looking for the callback 
     for(var i=arguments.length-1; i>=0; i--){ 
      (function(idx){ 

       //is this the function? 
       if(typeof arguments[idx] == 'function'){ 

        //keep a reference to the callback 
        var func = arguments[idx]; 

        //replace the callback 
        arguments[idx] = function(){ 
         try{ 
          //call the original callback 
          func.apply(self, arguments); 
         }catch(e){ 
         } 
        }; 
       } 
      })(i) 
     } 

     //call the original `on` with our new arguments 
     return _on.apply(this, arguments); 
    }; 
} 

演示：Fiddle