错误：类型“{}”缺少类型中的以下属性

Question

interface Person {
  name: string;
  surname: string;
}
let person1: Person = {};

person1.name = "name"
person1.surname = "surname"

When I declare person1 I get this error:

Type '{}' is missing the following properties from type Person

Answer 1

This is a better way:

let person1: Person = {name: '', surname: ''};

But if you want exactly empty object than you can hack it like this:

let person1: Person = {} as Person;

Update after comment:

Look at this unpredictableFunction :

const unpredictableFunction = (): string|number|string[] => {
  return Math.random() > 0.5 ? 'string' : Math.random() > 0.5 ? 9999 : ['1', '2', '3']
};

It may return number or it may return string or it may return array of strings

const person: Person = {name: '', surname: ''};
person.name = unpredictableFunction (); // this is a case you are talking about

In this case you will see

Type 'string | number | string[]' is not assignable to type 'string'.

Answers are:

Look at your code and ensure that you assign only strings to a Person properties,

Or update interface to be ready to a different values:

interface Person {
  name: string | number | string[];
  surname: string; 
}

Answer 2

You have defined an interface with two required properties. So when you define an object with the type of the Person interface you must define these properties right away like this:

let person: Person = {
    name: '',
    surname: ''
}

However if you believe these properties are not required but are rather optional you can change your interface to this:

interface Person {
    name?: string;
    surname?: string;
}

Using the ? syntax you mark the property as optional. The following code should then work:

let person: Person = {};

Answer 3

in Typescript 2.0 we can do this better

let person1 ! : Person;

this "!" is Non-null assertion operator

according to documentation

A new ! post-fix expression operator may be used to assert that its operand is non-null and non-undefined in contexts where the type checker is unable to conclude that fact. Specifically, the operation x! produces a value of the type of x with null and undefined excluded. Similar to type assertions of the forms x and x as T, the ! non-null assertion operator is simply removed in the emitted JavaScript code.