JS与JQ合并对象

如果需要将下面的json对象合并：

var a ={"a":"1","b":"2"}

var b ={"c":"3","d":"4","e":"5"}

想得到结果：

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

直接使用js的话，可以用一下方法：

<script>

function extend(des, src, override){

if(src instanceof Array){

for(var i = 0, len = src.length; i < len; i++)

extend(des, src[i], override);

}

for( var i in src){

if(override || !(i in des)){

des[i] = src[i];

}

}

return des;

}

var a ={"a":"1","b":"2"}

var b ={"c":"3","d":"4","e":"5"}

var c = extend({}, [a,b]);

alert(c.a);

</script>

当然如果你加载了jquery,那就更方便了。可以使用$.extend()方法，该方法有两种模式;

jQuery.extend( target [, object1 ] [, objectN ] ) #将后面的对象合并到新的对象中{}

jQuery.extend( [deep ], target, object1 [, objectN ] ) #将后面的对象合并到新的对象中{},如果深度deep为true则将递归合并成为新对象.

实例：

<script type="text/javascript">

# 请记得先加载jquery

var a ={"a":"1","b":"2"}

var b ={"c":"3","d":"4","e":"5"}

var c = $.extend({}, a,b);

console.log(c);

</script>
如果是：
var c = $.extend({}, [a,b]);
-->{"0":{"a":"1","b":"2"},"1":{"c":"3","d":"4","e":"5"}}