type MyProps = {
  workoutType: string | null;
};

JavaScript(和TypeScript)中的所有字段的值都可以是null或undefined。

您可以将字段设置为可选的，而不是可空的。

interface Employee1 {
    name: string;
    salary: number;
}

var a: Employee1 = { name: 'Bob', salary: 40000 }; // OK
var b: Employee1 = { name: 'Bob' }; // Not OK, you must have 'salary'
var c: Employee1 = { name: 'Bob', salary: undefined }; // OK
var d: Employee1 = { name: null, salary: undefined }; // OK

// OK
class SomeEmployeeA implements Employee1 {
    public name = 'Bob';
    public salary = 40000;
}

// Not OK: Must have 'salary'
class SomeEmployeeB implements Employee1 {
    public name: string;
}

比较:

interface Employee2 {
    name: string;
    salary?: number;
}

var a: Employee2 = { name: 'Bob', salary: 40000 }; // OK
var b: Employee2 = { name: 'Bob' }; // OK
var c: Employee2 = { name: 'Bob', salary: undefined }; // OK
var d: Employee2 = { name: null, salary: 'bob' }; // Not OK, salary must be a number

// OK, but doesn't make too much sense
class SomeEmployeeA implements Employee2 {
    public name = 'Bob';
}

就加个问号?到可选字段。

interface Employee{
   id: number;
   name: string;
   salary?: number;
}

type Nullable<T> = {
  [P in keyof T]: T[P] | null;
};

然后你就可以用它了

Nullable<Employee>

这样，您仍然可以像在其他地方一样使用Employee界面

我也有过同样的问题。ts中的所有类型都是可空的，因为void是所有类型的子类型(例如，与scala不同)。

看看这个流程图是否有帮助- https://github.com/bcherny/language-types-comparison#typescript

我通过编辑tsconfig解决了这个问题。json文件。

下面:"strict": true， 加上这两行:

"noImplicitAny": false,
"strictNullChecks": false,