CODEWITHSUNDEEP
typescript
prasanth
prasanth

Reputation: 311

Ignore certain properties in an interface while creating an object in typescript?

I have interface defined with some properties. I want to create an object from this interface, however while creating I don't want to fully initialize all the properties mentioned in the interface. I just want to initialize few. How can I achieve that? Thank you.

export interface Campaign {
  id: string;
  name: string;
  createdOn: string;
  lastUpdated: string;
  createdBy: User;
  type: CampaignType;
  status: CampaignStatus;
  startDate: string;
  endDate: string;
  budget: number;
  description: string;
  account: Account;
}

i want to create an array of campaign objects. this is what i am trying to do .

let campaigns: Campaign[] = [
    {  id:"1",
       name: "test campaign"
      }

   ];

however i get the following error .

Type '{ id: string; name: string; }' is missing the following properties from type 'Campaign': createdOn, lastUpdated, createdBy, type, and 6 more.ts(2740)

Upvotes: 9

Views: 9003

Answers (3)

Rannie Aguilar Peralta
Rannie Aguilar Peralta

Reputation: 1742

You could mark the properties that you don't want to set a value on creation as optional by doing { propName?: propType }

or if want to mark all the properties to be optional, you could use Partial

Partial<{
  prop1: string,
  prop2: number,
}>

Upvotes: 2

Naor Talmor
Naor Talmor

Reputation: 254

you can use Partial - 

let campaigns: Partial<Campaign>[] = [
    {  id:"1",
       name: "test campaign"
      }

   ];

Upvotes: 10

Z .
Z .

Reputation: 12837

if you make the optional properties nullable it should work.

example:

type User = {
  firstName: string;
  lastName?: string;
};

// We can assign a string to the "lastName" property
let john: User = { firstName: "John", lastName: "Doe" };

// ... or we can explicitly assign the value undefined
let jane: User = { firstName: "Jane", lastName: undefined };

// ... or we can not define the property at all
let jake: User = { firstName: "Jake" };

Upvotes: 7

Related Questions