Array as case in switch case JavaScript

Question

How can I use an array as case parameter in switch case?

switch ("value")
  case ArrayOfStrings // check every array item to be match with value
    ...

Answer 1

No.

This

switch ("value") {

  case ArrayOfStrings // check every array item to be match with value
    ...

does not work, because the value of switch and the value of case is checked with a Identity/strict equality operator === comparison.

It is not possible to check a value agains a value of the array.

Any other construction, like

switch (true) {
  case ArrayOfStrings.includes("value"):  // check every array item to be match with value

would work, but if you have only to check a single value and not other constraints, you better take

if (ArrayOfStrings.includes("value")) {
    // ...
}

Answer 2

Well first I would suggest using if with something like Array.isArray :

if (Array.isArray(value)) {
  // check array items
}

But, if you really want/need to check by comparing to Array constructor:

const check = (value) => {
  switch (value.constructor) {
    case Array:
      console.log('array');
      break;

    // For example
    case Number:
      console.log('number');
      break;
  }
}

check(7); // 'number'
check([]); // 'array'

I would really discourage from using this approach though.

Answer 3

In theory, you could do that to distinguish between different constructors.

class ArrayOfStrings extends Array<string> {};
class ArrayOfNumbers extends Array<number> {};

function main(input: ArrayOfStrings | ArrayOfNumbers): void {
  switch (input.constructor) {
    case ArrayOfStrings:
      console.log('It was an array of strings!');
      break;
    default:
       console.log('It was something else');
  }
}

main(new ArrayOfStrings('foo', 'bar')); // Logs 'It was an array of strings!'
main(new ArrayOfNumbers(1, 2)); // Logs 'It was something else'

However, this method has limitations .

1. It doesn't work with subclassing.
2. input will be of the same type for each case — as opposed to being discriminated. See TypeScript Playground .
3. You need to create your arrays by using the new keyword.

It's better to use if statements combined with type guards instead.