Checking if object is undefined with variable does not work Typescript

Question

I want to access a key in an object, but of course before doing so I need to check if the object is defined first. This works if I reference the key with a string, but typescript yells if I reference it with a variable assigned with a string:

type Fruit = "apple" | "orange";

type State = {
  [id in string]?: {
      [fruit in Fruit]: {
          status: "ripe" | "rotten";
      };
  };
};

const state: State = {};

const id = "hello";

if (state[id]) {
  state[id].apple.status = "ripe";
  ^^^^^^^^^
  Object is possibly 'undefined'.(2532)
}

if (state["hello"]) {
  // This is perfectly OK
  state["hello"].apple.status = "ripe";
}

Why is this?

See typescript playground

Yes, these checks don't work in TS. You need to check a variable, e.g., const myThing = state[id]; if (myThing) { myThing.apple.status = "ripe"; } — VLAZ
– VLAZ, Commented Oct 16, 2021 at 9:03

niko · Accepted Answer · 2021-10-16 09:02:35Z

1

You can use ! typescript ignores the fact that the value could be undefined which in your case is not

state[hello]!.apple.status = "ripe";

answered Oct 16, 2021 at 9:02

niko

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Comments

Mrudula Nudurupati · Accepted Answer · 2021-10-16 09:25:09Z

1

Use any of these instead of checking with just state['hello']

!!state['hello']
state['hello'] !== undefined

edited Oct 16, 2021 at 9:25

answered Oct 16, 2021 at 9:13

Mrudula Nudurupati

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3 Comments

VLAZ Over a year ago

"if (state['hello']) will always be true." not, if the property is missing, or the value is falsy then the if will not trigger.

VLAZ Over a year ago

if (!!state['hello']) is exactly the same as if (state['hello'])

VLAZ Over a year ago

Also, state['hello'] !== undefined (you have missing d at the end) will only work if the key is either missing or set to undefined, not if it's any other falsy value. Most notably null.

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