How do I dynamically assign properties to an object in TypeScript?
TypescriptTypescript Problem Overview
If I wanted to programatically assign a property to an object in Javascript, I would do it like this:
var obj = {};
obj.prop = "value";
But in TypeScript, this generates an error:
> The property 'prop' does not exist on value of type '{}'
How am I supposed to assign any new property to an object in TypeScript?
Typescript Solutions
Solution 1 - Typescript
Index types
It is possible to denote obj
as any
, but that defeats the whole purpose of using typescript. obj = {}
implies obj
is an Object
. Marking it as any
makes no sense. To accomplish the desired consistency an interface could be defined as follows.
interface LooseObject {
[key: string]: any
}
var obj: LooseObject = {};
OR to make it compact:
var obj: {[k: string]: any} = {};
LooseObject
can accept fields with any string as key and any
type as value.
obj.prop = "value";
obj.prop2 = 88;
The real elegance of this solution is that you can include typesafe fields in the interface.
interface MyType {
typesafeProp1?: number,
requiredProp1: string,
[key: string]: any
}
var obj: MyType ;
obj = { requiredProp1: "foo"}; // valid
obj = {} // error. 'requiredProp1' is missing
obj.typesafeProp1 = "bar" // error. typesafeProp1 should be a number
obj.prop = "value";
obj.prop2 = 88;
Record<Keys,Type>
utility type
> Update (August 2020): @transang brought this up in comments
> Record<Keys,Type>
is a Utility type in typescript. It is a much cleaner alternative for key-value pairs where property-names are not known.
> It's worth noting that Record<Keys,Type>
is a named alias to {[k: Keys]: Type}
where Keys
and Type
are generics.
> IMO, this makes it worth mentioning here
For comparison,
var obj: {[k: string]: any} = {};
becomes
var obj: Record<string,any> = {}
MyType
can now be defined by extending Record type
interface MyType extends Record<string,any> {
typesafeProp1?: number,
requiredProp1: string,
}
> While this answers the Original question, the answer here by @GreeneCreations might give another perspective on how to approach the problem.
Solution 2 - Typescript
Or all in one go:
var obj:any = {}
obj.prop = 5;
Solution 3 - Typescript
This solution is useful when your object has Specific Type. Like when obtaining the object to other source.
let user: User = new User();
(user as any).otherProperty = 'hello';
//user did not lose its type here.
Solution 4 - Typescript
I tend to put any
on the other side i.e. var foo:IFoo = <any>{};
So something like this is still typesafe:
interface IFoo{
bar:string;
baz:string;
boo:string;
}
// How I tend to intialize
var foo:IFoo = <any>{};
foo.bar = "asdf";
foo.baz = "boo";
foo.boo = "boo";
// the following is an error,
// so you haven't lost type safety
foo.bar = 123;
Alternatively you can mark these properties as optional:
interface IFoo{
bar?:string;
baz?:string;
boo?:string;
}
// Now your simple initialization works
var foo:IFoo = {};
Solution 5 - Typescript
Although the compiler complains it should still output it as you require. However, this will work.
const s = {};
s['prop'] = true;
Solution 6 - Typescript
I'm surprised that none of the answers reference Object.assign since that's the technique I use whenever I think about "composition" in JavaScript.
And it works as expected in TypeScript:
interface IExisting {
userName: string
}
interface INewStuff {
email: string
}
const existingObject: IExisting = {
userName: "jsmith"
}
const objectWithAllProps: IExisting & INewStuff = Object.assign({}, existingObject, {
email: "[email protected]"
})
console.log(objectWithAllProps.email); // [email protected]
Advantages
- type safety throughout because you don't need to use the
any
type at all - uses TypeScript's aggregate type (as denoted by the
&
when declaring the type ofobjectWithAllProps
), which clearly communicates that we're composing a new type on-the-fly (i.e. dynamically)
Things to be aware of
- Object.assign has it's own unique aspects (that are well known to most experienced JS devs) that should be considered when writing TypeScript.
- It can be used in a mutable fashion, or an immutable manner (I demonstrate the immutable way above, which means that
existingObject
stays untouched and therefore doesn't have anemail
property. For most functional-style programmers, that's a good thing since the result is the only new change). - Object.assign works the best when you have flatter objects. If you are combining two nested objects that contain nullable properties, you can end up overwriting truthy values with undefined. If you watch out for the order of the Object.assign arguments, you should be fine.
Solution 7 - Typescript
One more option do to that is to access the property as a collection:
var obj = {};
obj['prop'] = "value";
Solution 8 - Typescript
You can create new object based on the old object using the spread operator
interface MyObject {
prop1: string;
}
const myObj: MyObject = {
prop1: 'foo',
}
const newObj = {
...myObj,
prop2: 'bar',
}
console.log(newObj.prop2); // 'bar'
TypeScript will infer all the fields of the original object and VSCode will do autocompletion, etc.
Solution 9 - Typescript
Case 1:
var car = {type: "BMW", model: "i8", color: "white"};
car['owner'] = "ibrahim"; // You can add a property:
Case 2:
var car:any = {type: "BMW", model: "i8", color: "white"};
car.owner = "ibrahim"; // You can set a property: use any type
Solution 10 - Typescript
you can use this :
this.model = Object.assign(this.model, { newProp: 0 });
Solution 11 - Typescript
Since you cannot do this:
obj.prop = 'value';
If your TS compiler and your linter does not strict you, you can write this:
obj['prop'] = 'value';
If your TS compiler or linter is strict, another answer would be to typecast:
var obj = {};
obj = obj as unknown as { prop: string };
obj.prop = "value";
Solution 12 - Typescript
Simplest will be following
const obj = <any>{};
obj.prop1 = "value";
obj.prop2 = "another value"
Solution 13 - Typescript
Here is a special version of Object.assign
, that automatically adjusts the variable type with every property change. No need for additional variables, type assertions, explicit types or object copies:
function assign<T, U>(target: T, source: U): asserts target is T & U {
Object.assign(target, source)
}
const obj = {};
assign(obj, { prop1: "foo" })
// const obj now has type { prop1: string; }
obj.prop1 // string
assign(obj, { prop2: 42 })
// const obj now has type { prop1: string; prop2: number; }
obj.prop2 // number
// const obj: { prop1: "foo", prop2: 42 }
Note: The sample makes use of TS 3.7 assertion functions. The return type of assign
is void
, unlike Object.assign
.
Solution 14 - Typescript
To guarantee that the type is an Object
(i.e. key-value pairs), use:
const obj: {[x: string]: any} = {}
obj.prop = 'cool beans'
Solution 15 - Typescript
It is possible to add a member to an existing object by
- widening the type (read: extend/specialize the interface)
- cast the original object to the extended type
- add the member to the object
interface IEnhancedPromise<T> extends Promise<T> {
sayHello(): void;
}
const p = Promise.resolve("Peter");
const enhancedPromise = p as IEnhancedPromise<string>;
enhancedPromise.sayHello = () => enhancedPromise.then(value => console.info("Hello " + value));
// eventually prints "Hello Peter"
enhancedPromise.sayHello();
Solution 16 - Typescript
The best practice is use safe typing, I recommend you:
interface customObject extends MyObject {
newProp: string;
newProp2: number;
}
Solution 17 - Typescript
Store any new property on any kind of object by typecasting it to 'any':
var extend = <any>myObject;
extend.NewProperty = anotherObject;
Later on you can retrieve it by casting your extended object back to 'any':
var extendedObject = <any>myObject;
var anotherObject = <AnotherObjectType>extendedObject.NewProperty;
Solution 18 - Typescript
To preserve your previous type, temporary cast your object to any
var obj = {}
(<any>obj).prop = 5;
The new dynamic property will only be available when you use the cast:
var a = obj.prop; ==> Will generate a compiler error
var b = (<any>obj).prop; ==> Will assign 5 to b with no error;
Solution 19 - Typescript
Extending @jmvtrinidad solution for Angular,
When working with a already existing typed object, this is how to add new property.
let user: User = new User();
(user as any).otherProperty = 'hello';
//user did not lose its type here.
Now if you want to use otherProperty
in html side, this is what you'd need:
<div *ngIf="$any(user).otherProperty">
...
...
</div>
Angular compiler treats $any()
as a cast to the any
type just like in TypeScript when a <any>
or as any
cast is used.
Solution 20 - Typescript
dynamically assign properties to an object in TypeScript.
to do that You just need to use typescript interfaces like so:
interface IValue {
prop1: string;
prop2: string;
}
interface IType {
[code: string]: IValue;
}
you can use it like that
var obj: IType = {};
obj['code1'] = {
prop1: 'prop 1 value',
prop2: 'prop 2 value'
};
Solution 21 - Typescript
The only solution that is fully type-safe is this one, but is a little wordy and forces you to create multiple objects.
If you must create an empty object first, then pick one of these two solutions. Keep in mind that every time you use as
, you're losing safety.
Safer solution
The type of object
is safe inside getObject
, which means object.a
will be of type string | undefined
interface Example {
a: string;
b: number;
}
function getObject() {
const object: Partial<Example> = {};
object.a = 'one';
object.b = 1;
return object as Example;
}
Short solution
The type of object
is not safe inside getObject
, which means object.a
will be of type string
even before its assignment.
interface Example {
a: string;
b: number;
}
function getObject() {
const object = {} as Example;
object.a = 'one';
object.b = 1;
return object;
}
Solution 22 - Typescript
Late but, simple answer
`
let prop = 'name';
let value = 'sampath';
this.obj = {
...this.obj,
[prop]: value
};
`
Solution 23 - Typescript
If you are using Typescript, presumably you want to use the type safety; in which case naked Object and 'any' are counterindicated.
Better to not use Object or {}, but some named type; or you might be using an API with specific types, which you need extend with your own fields. I've found this to work:
class Given { ... } // API specified fields; or maybe it's just Object {}
interface PropAble extends Given {
props?: string; // you can cast any Given to this and set .props
// '?' indicates that the field is optional
}
let g:Given = getTheGivenObject();
(g as PropAble).props = "value for my new field";
// to avoid constantly casting:
let k:PropAble = getTheGivenObject();
k.props = "value for props";
Solution 24 - Typescript
You can add this declaration to silence the warnings.
declare var obj: any;
Solution 25 - Typescript
I ran into this problem when trying to do a partial update of an object that was acting as a storage for state.
type State = {
foo: string;
bar: string;
baz: string;
};
const newState = { foo: 'abc' };
if (someCondition) {
newState.bar = 'xyz'
}
setState(newState);
In this scenario, the best solution would be to use Partial<T>
. It makes all properties on the provided type optional using the ?
token. Read more about it in a more specific SO topic about making all properties on a type optional.
Here's how I solved it with Partial<T>
:
type State = {
foo: string;
bar: string;
baz: string;
};
const newState: Partial<State> = { foo: 'abc' };
if (someCondition) {
newState.bar = 'xyz';
}
setState(newState);
This is similar to what fregante described in their answer, but I wanted to paint a clearer picture for this specific use case (which is common in frontend applications).
Solution 26 - Typescript
I wrote an article tackling this very topic:
Typescript – Enhance an object and its type at runtime
https://tech.xriba.io/2022/03/24/typescript-enhance-an-object-and-its-type-at-runtime/
Maybe you could take inspiration from Typescript concepts such:
Solution 27 - Typescript
Use ES6 Map
whenever a map can take truly arbitrary values of fixed type, and optional properties otherwise
I think this is the guideline I'll go for. ES6 map can be done in typescript as mentioned at: https://stackoverflow.com/questions/30019542/es6-map-in-typescript
The main use case for optional properties are "options" parameters of functions: https://stackoverflow.com/questions/42108807/using-named-parameters-javascript-based-on-typescript/71679344#71679344 In that case, we do know in advance the exact list of allowed properties, so the sanest thing to do is to just define an explicit interface, and just make anything that is optional optional with ?
as mentioned at: https://stackoverflow.com/a/18444150/895245 to get as much type checking as possible:
const assert = require('assert')
interface myfuncOpts {
myInt: number,
myString?: string,
}
function myfunc({
myInt,
myString,
}: myfuncOpts) {
return `${myInt} ${myString}`
}
const opts: myfuncOpts = { myInt: 1 }
if (process.argv.length > 2) {
opts.myString = 'abc'
}
assert.strictEqual(
myfunc(opts),
'1 abc'
)
And then I'll use Map when it is something that is truly arbitrary (infinitely many possible keys) and of fixed type, e.g.:
const assert = require('assert')
const integerNames = new Map<number, string>([[1, 'one']])
integerNames.set(2, 'two')
assert.strictEqual(integerNames.get(1), 'one')
assert.strictEqual(integerNames.get(2), 'two')
Tested on:
"dependencies": {
"@types/node": "^16.11.13",
"typescript": "^4.5.4"
}
Solution 28 - Typescript
Try this:
export interface QueryParams {
page?: number,
limit?: number,
name?: string,
sort?: string,
direction?: string
}
Then use it
const query = {
name: 'abc'
}
query.page = 1