destructuring - 如何在javascript中解决这个问题?
问题描述
> Locate the `displayBirthdate` function you initially defined, which took no parameter. Modify it to use object de-structuring to get just the 'dob' property of the parameter object it will receive
here's the code
```javascript
const displayBirthdate = () => {};
const displayAddress = () => {};
const displayExtraUserInfo = (extra) => {
document.getElementById("btn-birthdate").addEventListener("click", ()=>
{ displayBirthdate(extra) })
document.getElementById("btn-phone").addEventListener("click", () =>
{ displayPhone(extra) })
document.getElementById("btn-address").addEventListener("click", () =>
{ displayAddress(extra) })
```
添加到上述问题,这是作为额外参数传递的预期响应
{
"results": [
{
"gender": "female",
"name": {
"title": "ms",
"first": "ceyhan",
"last": "dizdar"
},
"location": {
"street": "3826 şehitler cd",
"city": "tekirdağ",
"state": "bitlis",
"postcode": 11689,
"coordinates": {
"latitude": "79.1017",
"longitude": "27.1350"
},
"timezone": {
"offset": "+5:45",
"description": "Kathmandu"
}
},
"email": "ceyhan.dizdar@example.com",
"login": {
"uuid": "34eb65b2-0535-4656-bd68-4da69dc6d016",
"username": "orangefish864",
"password": "grandpa",
"salt": "vowzvAS2",
"md5": "cf4a7f3210ef97e8e72defafd80b94c8",
"sha1": "4f2af3439862b9bf25757ee73df8cd410ce201a2",
"sha256":
"1497acbca446b5fa47d4bc5ffe4e82c17818176596b66d94f213f091c8ed8077"
},
"dob": {
"date": "1979-08-10T22:03:55Z",
"age": 39
},
"registered": {
"date": "2008-05-24T13:30:20Z",
"age": 10
},
"phone": "(873)-801-4132",
"cell": "(751)-606-5317",
"id": {
"name": "",
"value": null
},
"picture": {
"large": "https://randomuser.me/api/portraits/women/59.jpg",
"medium": "https://randomuser.me/api/portraits/med/women/59.jpg",
"thumbnail":
"https://randomuser.me/api/portraits/thumb/women/59.jpg"
},
"nat": "TR"
}
],
"info": {
"seed": "008a9fe3a638239b",
"results": 1,
"page": 1,
"version": "1.2"
}
}
现在的问题是:编写一个箭头函数示例 displayBirthdate(),传入这个对象(额外)作为参数。使用解构方法,获取对象中的“dob”属性(额外)。以下是我试图解决这个问题的方法:
const displayBirthdate = (obj) =>{
const{results} = obj;
const[{dob}] = results;
}
但这似乎是不正确的。请任何帮助将不胜感激。谢谢你
解决方案
const displayBirthdate = ({dob}) => {};
const displayAddress ({location}) =>{};
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