Typescript property does not exist on type {}

Question

I have the following code in Typescript. Why does the compiler throws an error?

var object = {};
Object.defineProperty(object, 'first', {
     value: 37,
     writable: false,
     enumerable: true,
     configurable: true
});
console.log('first property: ' + object.first);

js.ts(14,42): error TS2339: Property 'first' does not exist on type '{}'.

It's the same code snippet like in the documentation of mozilla (examples section).

Why would you need Object.defineProperty at all? Whydont just make it `readonly` ? — Jonas Wilms, Mar 24 '18 at 17:04

Vayrex · Accepted Answer · 2018-03-24T16:44:09.923

9

Another way is to do interface, so compiler will know that property exists.

interface IFirst{
  first:number;
}


let object = {} as IFirst;
Object.defineProperty(object, 'first', {
  value: 37,
  writable: false,
  enumerable: true,
  configurable: true
});
console.log('first property: ' + object.first);

Take a look at this question How to customize properties in TypeScript

edited Mar 24 '18 at 16:44

answered Mar 24 '18 at 16:36

Vayrex

1,417
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It works fine. But is this the propper way to do it? ps: first should be of type number in your interface – Mar 24 '18 at 16:43
1

I think the line `let object = {} as IFirst;` will throw error cause the `first` property is required and missed in the type – Suren Srapyan Mar 24 '18 at 16:47
`const CONFIG_CACHE = {} as Config` works great... it's a shame such a small change isn't suggested in the hints. I now get t he reasoning of why `const CONFIG_CACHE: Config = {} ` doesn't work in hindsight. – Ray Foss Apr 19 '22 at 16:50

score 8 · Answer 2 · answered Nov 03 '18 at 16:30

8

Make the object type any:

var object: any = {};

answered Nov 03 '18 at 16:30

Emmanuel Osimosu

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Turning off type guards kinda defeats the purpose of using TypeScript. – mightimaus Jan 28 '21 at 17:35
strange answer, which I hope no one will use. Disabling typing usually leads to typing errors later on that you won't even notice. – Nikolay Feb 21 '23 at 11:47
there is no one size fit all... and i'm sure the OP can decide for themselves if they want the gaurd or not – Emmanuel Osimosu Mar 15 '23 at 09:57

score 0 · Answer 3 · answered Mar 24 '18 at 16:28

0

That's because Typescript is a strict type language. When you create a variable and give to it a type, you can't access properties that does not exists in that type. After adding extra property will not force the compiler to look for it. If you need to add a property after the creation, make the type of your variable any.

answered Mar 24 '18 at 16:28

Suren Srapyan

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2

Better to add an index signature `[k: string]: any` to the type instead of disabling all validation and all intellisense. – Aluan Haddad Mar 24 '18 at 20:26

Typescript property does not exist on type {}

3 Answers3

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