JS與JQ合併對象

若是須要將下面的json對象合併：

1. var a ={"a":"1","b":"2"}
   
2. var b ={"c":"3","d":"4","e":"5"}

想獲得結果：

1. var c ={"a":"1","b":"2","c":"3","d":"4","e":"5"}

直接使用js的話，能夠用一下方法：

1. <script>
   
2. function extend(des, src, override){
   
3.    if(src instanceof Array){
   
4.        for(var i = 0, len = src.length; i < len; i++)
   
5.             extend(des, src[i], override);
   
6.    }
   
7.    for( var i in src){
   
8.        if(override || !(i in des)){
   
9.            des[i] = src[i];
   
10.        }
    
11.    }
    
12.    return des;
    
13. }
    
14. var a ={"a":"1","b":"2"}
    
15. var b ={"c":"3","d":"4","e":"5"}
    
16. var c = extend({}, [a,b]);
    
17. alert(c.a);
    
18. </script>

固然若是你加載了jquery,那就更方便了。可使用$.extend()方法，該方法有兩種模式;

jQuery.extend( target [, object1 ] [, objectN ] )  #將後面的對象合併到新的對象中{}

jQuery.extend( [deep ], target, object1 [, objectN ] )  #將後面的對象合併到新的對象中{},若是深度deep爲true則將遞歸合併成爲新對象.

實例：

1. <script type="text/javascript">
   
2.     # 請記得先加載jquery
   
3.     var a ={"a":"1","b":"2"}
   
4.     var b ={"c":"3","d":"4","e":"5"}
   
5.     var c = $.extend({}, a,b);
   
6.     console.log(c);
   
7. </script>
   若是是：
   var c = $.extend({}, [a,b]);
   -->{"0":{"a":"1","b":"2"},"1":{"c":"3","d":"4","e":"5"}}