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Unable to 'index type' using a string variable

Time:04-12

I'm trying to invert an object's fields if a value exists, but I'm getting some error that I cannot decipher.

interface Control {
  name1: boolean;
  name2: boolean;
  ...
}

const values = ['name1', 'name2', ...];

const control: Control = {
  name1: false,
  name2: false,
  ...
}

for (const value of values) {
  if ((value in control) {
    control[value] = !control[value];  // error
  }
}

Error message:

Element implicitly has an 'any' type because expression of type 'string' can't be used to index type 'Control'.
  No index signature with a parameter of type 'string' was found on type 'Control'.ts(7053)

However, the error goes away if I explicitly pass one of the object's fields, such as control['name1']. Is there something I am not understanding?

CodePudding user response:

TypeScript doesn't know that your values array is an array of specific strings, only that it's a string[]. Try this:

interface Control {
  name1: boolean;
  name2: boolean;
}

const values = ['name1', 'name2'] as const;

const control: Control = {
  name1: false,
  name2: false,
}

for (const value of values) {
  if (value in control) {
    control[value] = !control[value];  // error
  }
}

TypeScript Playground

That as const tells TypeScript to specify the type of values more tightly to be ['name1', 'name2'] in this case, so when you're iterating through it value will be typed as 'name1' | 'name2' and TypeScript will understand that those values can be used to index a Control.

CodePudding user response:

The value variable can be any value during application runtime and TypeScript has no awareness that it is a valid key defined within the control object. Hence it cannot simply infer the value type of control[value].

The fix is however very simple, all that needs to be done is to tell TypeScript that the values array only contains valid keys from the Control type via the use of the keyof Control syntax:

const values: (keyof Control)[] = ['name1', 'name2'];

Here is a link to the Playground.

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