如何分配JSON对象

问题描述：

我是AngularJS中的新成员。我试图分配JSON对象，我使用Object.assign来分配JSON对象。但是我不能得到我想要的结果请帮我拿到结果。如何分配JSON对象

'use strict' 
 

 
let one ={"Start":"11","Dashboard":"1","Settings":"2","Support":"3","Trunk":"4","Routing":"5","Solutions":"6","Accounting":"7","Statistic":"8","Marketing":"9","Profile":"10"}; 
 
let two = {"Settings":{"Basic Settings":"1","Number Range":"2","Employess":"3","Message Fulfillment":"4"},"Support":{"All Contacts":"5","Detail Approval":"6","Incomplete Signup":"7","Settings":"8"},"Trunk":{"My Tickets":"9","All Tickets":"10"}}; 
 

 
let three \t = \t {"Settings":{"Basic Settings":{"My Information":"1","Bank Information":"2"}}}; 
 

 
let four = Object.assign({}, one, two, three); 
 

 
console.log(JSON.stringify(four));

我需要这样的： - { “开始”： “11”， “仪表板”： “1”， “设置”：{ “基本设置”：{ “我的信息”： “1”，”银行信息“：”2“}，”号码范围“：”2“，”职业“：”3“，”信息实现“：”4“}，”支持“：{”所有联系人“：”5“， “详细审批”：“6”，“未完成注册”：“7”，“设置”：“8”}，“中继线”：{“我的门票”：“9”，“所有门票”：“10”} ， “路由”： “5”， “解决方案”： “6”， “会计”： “7”， “统计”： “8”， “市场”： “9”， “个人资料”： “10”，”基本设置“：{”我的信息“：”1“，”银行信息“：”2“}} –

你有答案吗？ –

@deepgagan你检查了我的答案吗？ –

答

您可以使用jQuery $.extend(a, b);扩展对象，但因为使用的是angularjs，您可以使用JavaScript核心

let one ={"Start":"11","Dashboard":"1","Settings":"2","Support":"3","Trunk":"4","Routing":"5","Solutions":"6","Accounting":"7","Statistic":"8","Marketing":"9","Profile":"10"}; 
let two = {"Settings":{"Basic Settings":"1","Number Range":"2","Employess":"3","Message Fulfillment":"4"},"Support":{"All Contacts":"5","Detail Approval":"6","Incomplete Signup":"7","Settings":"8"},"Trunk":{"My Tickets":"9","All Tickets":"10"}}; 

let three = {"Settings":{"Basic Settings":{"My Information":"1","Bank Information":"2"}}}; 


function jsonExtend(a, b) { 
for (var key in b) { 
    a[key] = b[key]; 
} 
return a; 
} 

var output = {}; 
output = jsonExtend(output, one); 
output = jsonExtend(output, two); 
output = jsonExtend(output, three); 

console.log(output)

输出不显示： - “Number Range”：“2”，“Employess”：“3”，“Message Fulfillment”：“4” –

当连接对象时，属性被重复，因此第三个覆盖第二个。 – vusan

好的。但我需要这些值。那么我怎么能得到这些值？ –

答

试试这个：

function mergeObjects() { 
 
    var merged_obj = {}; 
 
    for (i in arguments) { 
 
     obj = arguments[i]; 
 
     for (j in obj) { 
 
       merged_obj[j] = obj[j]; 
 
     } 
 
    } 
 
    return merged_obj; 
 
} 
 

 
var one ={"Start":"11","Dashboard":"1","Settings":"2","Support":"3","Trunk":"4","Routing":"5","Solutions":"6","Accounting":"7","Statistic":"8","Marketing":"9","Profile":"10"}; 
 
var two = {"Settings":{"Basic Settings":"1","Number Range":"2","Employess":"3","Message Fulfillment":"4"},"Support":{"All Contacts":"5","Detail Approval":"6","Incomplete Signup":"7","Settings":"8"},"Trunk":{"My Tickets":"9","All Tickets":"10"}}; 
 

 
var three = {"Settings":{"Basic Settings":{"My Information":"1","Bank Information":"2"}}}; 
 

 
var mergedObj = mergeObjects(one, two, three); 
 
console.log(mergedObj);

它不显示这些值： - “编号范围”：“2”，“职业”：“3”，“消息实现”：“4” –

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