就像标题说的那样。我想从下面数组中的数据创建一个新对象: 对象的名称是名字和姓氏。
//this is the input
arrayToObject([['Christ', 'Evans', 'Male'], ['Robert', 'Downey', 'Male']]);
//this is output that i want
// 1. Christ Evans:
// { firstName: 'Christ',
// lastName: 'Evans',
// gender: 'Male',
// 2. Robert Downey:
// { firstName: 'Robert',
// lastName: 'Downey',
// gender: 'Male',
// this is code that i write
function arrayToObject(arr) {
let ObjName=""
for (let i = 0; i<arr.length;i++){
let firstName=arr[i][0]
let lastName=arr[i][1]
let ObjName=firstName.concat(' ',lastName)
let gender=arr[1][2]
ObjName = { // idk how to make the object.
'firstName':firstName,
'lastName':lastName,
'gender':gender,
}
}
}
我在我可以声明该对象的部分中苦苦挣扎。
答案 0 :(得分:0)
您可以使用Object.fromEntries
的嵌套方法,并使用keys数组映射属性。
#4
var data = [['Christ', 'Evans', 'Male'], ['Robert', 'Downey', 'Male']],
keys = ['firstName', 'lastName', 'gender'],
result = Object.fromEntries(data.map(a => [
a.slice(0, 2).join(' '),
Object.fromEntries(keys.map((k, i) => [k, a[i]]))
]));
console.log(result);
答案 1 :(得分:-1)
我想您想要一个具有以下结构的对象数组:
[
{
"firstName": "Christ",
"lastName": "Evans",
"gender": "Male"
},
{
"firstName": "Robert",
"lastName": "Downey",
"gender": "Male"
}
]
在这种情况下,您可以简单地使用Array.prototype.map
,并且在回调中只需使用数组解构将第二级数组解压缩为firstName
,lastName
和{{1} }:
gender