#!/usr/bin/env python # encoding: utf-8 import json def read(obj,key): collect = list() for k in obj: v = obj[k] if isinstance(v,dict): collect.extend(read(v,k)) elif isinstance(v,list): if key=='': collect.extend(readList(v,k)) else: collect.extend(
已知一个多层嵌套的json,取出所有父元素和子元素的id值 思路:因为不知道到底嵌套了多少层,递归有可能造成栈溢出.查询时间特别久的问题 所以先查询一次,判断是否有子节点,如果有,取出子节点并到父节点上,并动态更改数据长度,这样无限循环处理json取出所有id menuIdInit () { var _this = this; var _menu = _this.menus; var menuId = []; var len = _menu.length; for(var i = 0; i <
如题所示,代码如下: var arJsonNesting = [{id:1,name:"zhang3" ,children:[{id:2,name:"zhang33"},{id:3,name:"zhang44"}]} ,{id:4,name:"li4"}];//注意有的JSON对象有子对象children var arJson = new Array(); function refining(arJsonNesting,arJ
Newtonsoft.Json.Net20.dll 下载请访问http://files.cnblogs.com/hualei/Newtonsoft.Json.Net20.rar 在.net 2.0中提取这样的json {"name":"lily","age":23,"addr":{"city":guangzhou,"province":guangdong}} 引用命名空间 using N
var map = new Array();//二维数组 var map2 = new Array();//一维数组 for (var i = 0; i < e.Data.length; i++) { map[i]= {value:e.Data[i].market_value,name:e.Data[i].name};//二维数组 map2.push(e.Data[i].name);//一维数组 } console.log(JSON.stringify(map)); console.log(JS