javascript Ajax method in different browser has different result

Question

I have a code like this :

var width = [];

$.ajax({
   url : "http://xxx.xxx.xxx" ,
   type : "Get",
   dataType: "json",
   async   : true,
   success : function(data){
       for (var index in data)
       {
           var options = data[index].option;
           var position = (++index);
           width[index] = new Array();
           for (var keys in options)
           {
              width[index][keys] = options[keys].votes;
           }
       }
   }
});

in the chrome browser, it works fine, but in the safari, the error look likes:

TypeError: undefined is not an object (evaluating 'width[index][keys] = options[keys].votes')

But if I open debug mode in safari, there is no error, so what's the problem in this code?

Are you saying that the error disappears if you open the debug tools in safari? Or just that there's no other additional error being reported? — James Thorpe, May 09 '16 at 15:14
Once, you use jQuery why do you mix it with pure javascrtipt? Just a question. — vaso123, May 09 '16 at 15:15
@lolka_bolka jQuery is JavaScript. Why would you replace any of their "pure JavaScript" stuff with jQuery helpers? All they're using is a couple of for-in loops. — Mike Cluck, May 09 '16 at 15:17
Probably you've got some polyfills in Safari? See [How to define method in javascript on Array.prototype and Object.prototype so that it doesn't appear in for in loop](http://stackoverflow.com/q/13296340/1048572) — Bergi, May 09 '16 at 15:21
Start off with iterating arrays with a regular for loop ([no for..in](http://stackoverflow.com/questions/500504/why-is-using-for-in-with-array-iteration-a-bad-idea)). — A1rPun, May 09 '16 at 15:21
If i open debug tools in safari, thers is no error and no other additional error being reported, but If I close the debug tools and refresh, the error will appears. — linbob, May 09 '16 at 15:23
You're incrementing the loop counter inside the loop (`++index`). Don't do that. — Reinstate Monica Cellio, May 09 '16 at 15:44
As @Archer pointed out - you are incrementing your loop variable `index` while using it inside the loop. — trenthaynes, May 09 '16 at 19:27

score 0 · Accepted Answer · answered May 10 '16 at 08:13

I change my for loop code like this:

    for (var index = 0;index < data.length;index++) {
        var options = data[index].option;
        var position = index + 1;
        width[index] = new Array();
        for (var keys = 0; keys < options.length; keys++) {
            width[index][keys] = options[keys].votes;
        }
    }

there is no error in any browser.

javascript Ajax method in different browser has different result

1 Answers1