如何分配JSON对象

Question

我是AngularJS的新手。 我正在尝试分配JSON对象，并且我正在使用Object.assign分配JSON对象。 但是我无法获得想要的结果，请帮助我获得结果。

 '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 = {"Settings":{"Basic Settings":{"My Information":"1","Bank Information":"2"}}}; let four = Object.assign({}, one, two, three); console.log(JSON.stringify(four)); 

Answer 1

尝试这个 ：

 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); 

Answer 2

您可以使用$.extend(a, b);扩展对象$.extend(a, b); 使用jQuery，但是当您使用angularjs时，可以使用核心javascript

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)