如何分配JSON對象

我是AngularJS中的新成員。我試圖分配JSON對象，我使用Object.assign來分配JSON對象。但是我不能得到我想要的結果請幫我拿到結果。如何分配JSON對象

'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));

來源

2017-04-06 deep gagan

我需要這樣的： - { 「開始」：「11」，「儀表板」：「1」，「設置」：{ 「基本設置」：{ 「我的信息」：「1」，」銀行信息「：」2「}，」號碼範圍「：」2「，」職業「：」3「，」信息實現「：」4「}，」支持「：{」所有聯繫人「：」5「，「詳細審批」：「6」，「未完成註冊」：「7」，「設置」：「8」}，「中繼線」：{「我的門票」：「9」，「所有門票」：「10」} ，「路由」：「5」，「解決方案」：「6」，「會計」：「7」，「統計」：「8」，「市場」：「9」，「個人資料」：「10」，」基本設置「：{」我的信息「：」1「，」銀行信息「：」2「}} –

你有答案嗎？ –

@deepgagan你檢查了我的答案嗎？ –

您可以使用jQuery $.extend(a, b);擴展對象，但因爲使用的是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)

來源

2017-04-06 11:22:55 vusan

輸出不顯示： - 「Number Range」：「2」，「Employess」：「3」，「Message Fulfillment」：「4」 –

當連接對象時，屬性被重複，因此第三個覆蓋第二個。 – vusan

好的。但我需要這些值。那麼我怎麼能得到這些值？ –

試試這個：

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);

來源

2017-04-06 12:44:49

它不顯示這些值： - 「編號範圍」：「2」，「職業」：「3」，「消息實現」：「4」 –

如何分配JSON對象

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