I have an interface in TypeScript.
interface Employee{
id: number;
name: string;
salary: number;
}
I would like to make salary as a nullable field (Like we can do in C#). Is this possible to do in TypeScript?
I have an interface in TypeScript.
interface Employee{
id: number;
name: string;
salary: number;
}
I would like to make salary as a nullable field (Like we can do in C#). Is this possible to do in TypeScript?
All fields in JavaScript (and in TypeScript) can have the value null or undefined.
You can make the field optional which is different from nullable.
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;
}
Compare with:
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';
}
Union type is in my mind best option in this case:
interface Employee{
id: number;
name: string;
salary: number | null;
}
// Both cases are valid
let employe1: Employee = { id: 1, name: 'John', salary: 100 };
let employe2: Employee = { id: 1, name: 'John', salary: null };
EDIT : For this to work as expected, you should enable the strictNullChecks in tsconfig.
To be more C# like, define the Nullable type like this:
type Nullable<T> = T | null;
interface Employee{
id: number;
name: string;
salary: Nullable<number>;
}
Bonus:
To make Nullable behave like a built in Typescript type, define it in a global.d.ts definition file in the root source folder. This path worked for me: /src/global.d.ts
You can just implement a user-defined type like the following:
type Nullable<T> = T | undefined | null;
var foo: Nullable<number> = 10; // ok
var bar: Nullable<number> = true; // type 'true' is not assignable to type 'Nullable<number>'
var baz: Nullable<number> = null; // ok
var arr1: Nullable<Array<number>> = [1,2]; // ok
var obj: Nullable<Object> = {}; // ok
// Type 'number[]' is not assignable to type 'string[]'.
// Type 'number' is not assignable to type 'string'
var arr2: Nullable<Array<string>> = [1,2];
Nullable type can invoke runtime error.
So I think it's good to use a compiler option --strictNullChecks and declare number | null as type. also in case of nested function, although input type is null, compiler can not know what it could break, so I recommend use !(exclamination mark).
function broken(name: string | null): string {
function postfix(epithet: string) {
return name.charAt(0) + '. the ' + epithet; // error, 'name' is possibly null
}
name = name || "Bob";
return postfix("great");
}
function fixed(name: string | null): string {
function postfix(epithet: string) {
return name!.charAt(0) + '. the ' + epithet; // ok
}
name = name || "Bob";
return postfix("great");
}
Reference. https://www.typescriptlang.org/docs/handbook/advanced-types.html#type-guards-and-type-assertions
i had this same question a while back.. all types in ts are nullable, because void is a subtype of all types (unlike, for example, scala).
see if this flowchart helps - https://github.com/bcherny/language-types-comparison#typescript