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>
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» js和jquery中json对象合并的方法