Question

我创建一个JSON对象并将其存储到mysql然后再次检索并解析它，当我对JSON对象进行字符串化时，它将属性置于“”之间，当我尝试解析检索到的字符串时，我得到意外的输入结束，这是代码

for(var i=0;i<objects.length;i++){
        var o=new Object();
        var id=objects[i].userData.id;
        var name=objects[i].name;
        o.id=id;
        o.x=objects[i].position.x;
        o.y=objects[i].position.y;
        o.z=objects[i].position.z;
        o.r=objects[i].rotation.z;
        resources[name]=o;
}

我将JSON.stringify(resources)插入mysql，插入的字符串如{"animalsherd0":{"id":"11","x":"4.7","y":"19.6","z":"18.8","r":0},"oasis1":{"id":"19","x":"-11.3","y":"19.6","z":"18.8","r":0},"corn2":{"id":"24","x":"-5.6","y":"19.6","z":"5.0","r":0}}

检索到的字符串存储在变量中，那我该如何解析这个字符串呢？

Answer 1

试试这个： jQuery.parseJSON

var obj = jQuery.parseJSON('{"animalsherd0":{"id":"11","x":"4.7","y":"19.6","z":"18.8","r":0},"oasis1":{"id":"19","x":"-11.3","y":"19.6","z":"18.8","r":0},"corn2":{"id":"24","x":"-5.6","y":"19.6","z":"5.0","r":0}}');
console.log(obj);

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>

parse字符串包含从mysql检索到JSON的'''

1 个答案: