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"this.users = users;" not binding to "users;" property(“this.users=用户；”不绑定到“用户；”属性)

转载 作者：bug小助手 更新时间：2023-10-22 16:38:52 26 4
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export class UserListComponent implements OnInit{
  users;

  constructor(private userService: UserService) {}

  ngOnInit(){
    this.userService.getUsers()
    .subscribe(users=> {
      console.log(users);
      this.users = users;
    });
  }
}

The "users;" property above displays a
"Member 'users' implicitly has an 'any' type.ts(7008)
(property) UserListComponent.users: any"
error. help me

上面的“users；”属性显示“成员“users”隐式具有“any”类型。ts（7008）（property）UserListComponent.users:any”错误。帮帮我


This is an angular CLI project leveraging the github "https://api.github.com/users" api.

这是一个利用github的有角度的CLI项目“https://api.github.com/users“api。


更多回答

So what is the type of users? It should be something like users: UserModel[]; or similar. Worst case, you could do users: any;, but any should be avoided.

那么用户的类型是什么呢？它应该类似于users:UserModel[]；或类似的。最坏的情况是，你可以做用户：任何；，但是应该避免任何情况。

优秀答案推荐

The quick fix is to change users; to users: any;, if you don't care about typings.

快速解决办法是更换用户；对用户：任何；，如果你不在乎打字员的话。


If you do care about typings you should type users the same as the return value of getUsers.subscribe().

如果您确实关心键入，那么您应该键入与getUsers.subscribe（）的返回值相同的user。


Assuming getUsers() returns something along the lines of Observable<User[]>,
your initial declaration of users could be users: User[] = []; or users: User[] | null = null;

假设getUsers（）返回的内容与Observable＜User[]＞类似，那么用户的初始声明可能是users:User[]＝[]；或users:User[]| null=null；



this.userService.getUsers()
  subscribe((users:any) => {
  console.log(users);
  this.users = users;
});

Modify like this while subscribing data from observable.

在订阅observable的数据时这样修改。


Add any (users:any) as type for incoming response from observable.

添加any（users:any）作为来自observable的传入响应的类型。



When TypeScript is unable to infer a variables/members type because there are no type annotations present it sometimes falls back to the type any. This compiler option is enabled per default.

当TypeScript由于没有类型注释而无法推断变量/成员类型时，它有时会返回到类型any。默认情况下启用此编译器选项。


To prevent this error:

要防止此错误，请执行以下操作：



  • Set set an explicit Type Annotation for users. users is likely going to be something like User[].

  • Disable the noImplicitAny compiler option. (Not recommended)



Change users; to

更改用户；到


users: any[]=[];

users:any[]=[]；


or to

或


users: any;

用户：任意；


更多回答

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