[英]How to assign JSON object
我是AngularJS的新手。 我正在嘗試分配JSON對象,並且我正在使用Object.assign分配JSON對象。 但是我無法獲得想要的結果,請幫助我獲得結果。
'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));
嘗試這個 :
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);
您可以使用$.extend(a, b);
擴展對象$.extend(a, b);
使用jQuery,但是當您使用angularjs時,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)
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