Question

我有json数据，我想将其转换为对象格式以进行创建操作。

json

[
{
    "user": {
        "id": 83,
        "username": "das",
        "first_name": "dsafha",
        "last_name": "dfksdfk",
        "email": "sasda@gmail.com",
        "is_active": true,
        "is_superuser": false
    },
    "role": "testBu"
},
{
    "user": {
        "id": 84,
        "username": "sadfds",
        "first_name": "dshhgds",
        "last_name": "fsdjsl",
        "email": "fdjgd@gmail.com",
        "is_active": true,
        "is_superuser": false
    },
    "role": "testeditrole"
},
{
    "user": {
        "id": 86,
        "username": "fs",
        "first_name": "efhks",
        "last_name": "sofdh",
        "email": "fdshk@gmail.com",
        "is_active": true,
        "is_superuser": false
    },
    "role": "testeditrole"
},
{
    "user": {
        "id": 87,
        "username": "xz",
        "first_name": "vj",
        "last_name": "vkfd",
        "email": "sdsl@gmail.com",
        "is_active": true,
        "is_superuser": false
    },
    "role": "testeditrole"
}
]

我试图这样做

componet.ts

let user:any = {};
   user["username"] = this.user.user.email
   user["first_name"] = this.user.user.first_name
   user["last_name"]= this.user.user.last_name
   user["email"]= this.user.user.email

   this.userData["user"] = user
   this.userData["role"] = this.user.role

并且在将数据输入到输入字段后感到安慰时，我没有获得角色数据。我确实得到了其他数据。角色数据显示为空。我确实认为这是因为我是如何在componet.ts中编写代码的原因。

Answer 1

您始终可以创建模型：

export class User {
  id: string,
  username: string,
  first_name: string,
  last_name: string,
  email: string,
  is_active: boolean,
  is_superuser: boolean
}

export class UserDetails{
  user:User;
  role:string;
}

然后

// Assume you have received the json in string form in 'resultlist' variable
let dataList = <Array<UserDetails>>JSON.parse(resultlist);

OR

// Assume you have received the json in object form in 'resultlist' variable
let dataList = <Array<UserDetails>>resultlist;

快乐编码！

Answer 2

我认为最好的方法是，创建一个与JSON结构相对应的对象，然后仅将数据JSON分配给对象的数组。

class User{
   id:string;
   username:string;
   firstName:string;
   lastName:string;
   email:string;
   isActive:boolean;
   isSuperviser:boolean;
}
class JSONData{
   user:User;
   role:string
}
 data:JSONData[] = yourJson.data;

您可以像在类中一样调整json数据中的属性名称，并且可以正确，轻松地处理数据，请在此链接上查阅more infos

Answer 3

this.userData["role"] = this.user.role应该是this.userData["role"] = this.role

Answer 4

无需转换。它已经是json格式，但是您应该使用数组索引，然后设置用户变量。

例如：

let user:any = {};
user = this.user[arrayIndex].user;

this.userData["user"] = user
this.userData["role"] = this.user[arrayIndex].role;

如果要从json数组获取数据，则应使用数组索引。这样。

let user:any = {};
user = this.user[0].user;
this.userData["role"] = this.user[0].role;

将json数据转换为component.ts中的对象

4 个答案: