js中的JSON数据为什么取不出来?

2024-12-01 13:39:52
推荐回答(1个)
回答1:

var json = { contry:{ area:{ man:"12万", women:"10万" } } };
//方式一:使用eval解析
var obj = eval(json);
alert(obj.constructor);
alert(obj.contry.area.women);

//方式二:使用Funtion函数
var strJSON = "{name:'json name'}";//得到的JSON
var obj = new Function("return" + strJSON)();//转换后的JSON对象
alert(obj.name);//json name
alert(obj.constructor);

//复杂一点的json数组数据的解析
var value1 = [{"c01":"1","c02":"2","c03":"3","c04":"4","c05":"5","c06":"6","c07":"7","c08":"8","c09":"9"}, {"c01":"2","c02":"4","c03":"5","c04":"2","c05":"8","c06":"11","c07":"21","c08":"1","c09":"12"}, {"c01":"5","c02":"1","c03":"4","c04":"11","c05":"9","c06":"8","c07":"1","c08":"8","c09":"2"}]; var obj1 = eval(value1);
alert(obj1[0].c01);

//复杂一点的json的另一种形式
var value2 = {"list":[ {"password":"1230","username":"coolcooldool"}, {"password":"thisis2","username":"okokok"}], "array":[{"password":"1230","username":"coolcooldool"},{"password":"thisis2","username":"okokok"}]};

var obj2 = eval(value2);
alert(obj2.list[0].password);