使用firebug调试jquery或javascript。-Debuggingjqueryorjavascriptusingfirebug

I wrote a code in jquery. I was not running initially, then i checked online jslint for syntax errors. I caught some errors. Now still the code was not working as expected. So i went for firebug. I haven't done a lot of debugging. I am new to it. Here is my code

我在jquery中编写了一个代码。我最初没有运行，然后检查了在线jslint的语法错误。我发现一些错误。现在，代码仍然没有达到预期的效果。所以我去找firebug。我没有做过很多调试。我是新手。这是我的代码

var j = 2;
var friends = [];
var distance =[];


$(document).ready(function () {

      $('#button').click(function () {
        if (j <11) {
            $('#friends').append('Friend' + j + ':

');
            j++;
        }
        else {
            alert("Limit reached");
        }
    });



   $('button').click(function(){
  console.log("button clicked");
   var a =[];
    for(i=1;i<=j;i++)
     {
        a[i] = $("#friend" + i).val();
      }     


    var gurl = "http://maps.googleapis.com/maps/api/distancematrix/json?"+
        "origins=" +
        a.join('|').replace(/ /g,'+') +
        "&destinatiOns=" +
        a.join('|').replace(/ /g,'+') +
        "&sensor=false";


  jQuery.ajax(
          {
             type: "GET",
             url: gurl,
             dataType: 'jsonp'
          }).done(function (response) 
             {
                var rows = response.rows;
                  alert("hello there");

                for (var i = 0; i 

 

    

    
    
    
    
    
    


 

  
    
        
        Your Place:

        


        

         Friend1:

        

        

        

            Add!
        

        


        GO!

        


        



        
    

  

console.log("distance : ");
console.log(distance);