Object.keys.map 不返回原始键

Question

我正在从我的数据库中获取一个对象： 示例：

{availability: null
bio: null
category: ""
createdAt: "2020-10-13T13:47:29.495Z"
email: "oznerol@prova.com"
firstName: "Lorenzo"
id: 37
image: null
job: null
lastName: "Del Rosario"
location: ""
password: "$2b$10$J6m71nc8gBXTvEENMwBjt.azNitCsna3LsxaDIUg.B0ztiTm5xFea"
phone: ""
skills: null
updatedAt: "2020-10-13T13:47:29.495Z"
website: ""}

我正在尝试映射此对象以将所有 null 值更改为 ''。

这是我的代码：

const [userData, setUserData] = useState()

useEffect( ()=>{
    
  const fetchData = async () => {
    const a = await getUserData(userId);//this retuns the object
    const cleanUserData = Object.entries(a).map(([key, value]) =>
      value === null || undefined ? value = '' : value
    )
    setUserData(cleanUserData);
  };
  fetchData();
},[])

但是通过这种方式我改变了我的值，但我失去了我的钥匙，结果是一个像这样的对象：

{0:37
1:"oznerol@prova.com"
2:"$2b$10$J6m71nc8gBXTvEENMwBjt.azNitCsna3LsxaDIUg.B0ztiTm5xFea"
3:"Lorenzo"
4:"Del Rosario"
5:""
6:""
7:""
8:""
9:""
10:""
11:""
12
...}

Answer 1

Object.entries 和 Object.keys 都返回一个数组，因此当您使用 map 时，结果将是一个数组。使用reduce 来创建新对象，如下所示：

const cleanUserData = Object.entries(a).reduce((acc, [key, value]) => {
  acc[key] = value == null ? '' : value;
  return acc;
}, {});

注意：您的原始测试 value === null || undefined 不正确，因为这不是您针对两个可能值进行测试的方式，它应该是 value === null || value === undefined。测试value == null（使用双等号==）也足以与javascript中的undefined == null一样。

演示：

const a = {availability: null, bio: null, category: "", createdAt: "2020-10-13T13:47:29.495Z", email: "oznerol@prova.com", firstName: "Lorenzo", id: 37, image: null, job: null, lastName: "Del Rosario", location: "", password: "$2b$10$J6m71nc8gBXTvEENMwBjt.azNitCsna3LsxaDIUg.B0ztiTm5xFea", phone: "", skills: null, updatedAt: "2020-10-13T13:47:29.495Z", website: ""};

const cleanUserData = Object.entries(a).reduce((acc, [key, value]) => {
  acc[key] = value == null ? '' : value;
  return acc;
}, {});

console.log(cleanUserData);

Answer 2

您可能需要使用例如Object.fromEntries 映射后重新创建主对象。

const o = {availability: null, bio: null, category: '', createdAt: "2020-10-13T13:47:29.495Z", email: "oznerol@prova.com", firstName: "Lorenzo", id: 37, image: null, job: null, lastName: "Del Rosario", location: "", password: "$2b$10$J6m71nc8gBXTvEENMwBjt.azNitCsna3LsxaDIUg.B0ztiTm5xFea", phone: '', skills: null, updatedAt: "2020-10-13T13:47:29.495Z", website: "",};

const res = Object.fromEntries(
   Object.entries(o).map(([key, val]) => [key, val ?? '']),
);

console.log(res);
// then set the state -> setUserData(res);

Answer 3

请检查此解决方案

 Object.keys(x).reduce((acc, key)=> {
    acc[key] = x[key] === null ? '' : x[key];
    return acc;
    }, {})

附言

Object.entries() 方法返回给定对象自己的数组 可枚举的字符串键属性 [key, value] 对。

这就是为什么你会收到一个索引而不是真正的键

Answer 4

这里是通过复制对象的可能解决方案：

const data ={
  availability: null,
  bio: null,
  category: "",
  createdAt: "2020-10-13T13:47:29.495Z",
  email: "oznerol@prova.com",
  firstName: "Lorenzo",
  id: 37,
  image: null,
  job: null,
  lastName: "Del Rosario",
  location: "",
  password: "$2b$10$J6m71nc8gBXTvEENMwBjt.azNitCsna3LsxaDIUg.B0ztiTm5xFea",
  phone: "",
  skills: null,
  updatedAt: "2020-10-13T13:47:29.495Z",
  website: ""
}; 


let cleanUserData = (data) => {
    const cleanData = {};
    Object.keys(data).forEach((key) =>
      cleanData[key] = data[key] === null ? '' : data[key]
    )
    return cleanData;
  };

console.log(cleanUserData(data));

Answer 5

很简单，您可以使用 map reduce 函数来获取结果。

const obj = {
  "availability": null,
  "bio": null,
  "category": "",
  "createdAt": "2020-10-13T13:47:29.495Z",
  "email": "oznerol@prova.com",
  "firstName": "Lorenzo",
  "id": 37,
  "image": null,
  "job": null,
  "lastName": "Del Rosario",
  "location": "",
  "password": "$2b$10$J6m71nc8gBXTvEENMwBjt.azNitCsna3LsxaDIUg.B0ztiTm5xFea",
  "phone": "",
  "skills": null,
  "updatedAt": "2020-10-13T13:47:29.495Z",
  "website": ""
}

const result = Object.keys(obj).reduce((acc, curr) => {
    acc[curr] = (obj[curr] === null || obj[curr] === undefined)  ? '' : obj[curr]
    return acc;
}, {})

console.log(result)