Is it possible for me to create separate class/interface based on property values?

Question

Is it possible for me to create separate class/interface based on property values?

My question pertains to a class named Selection (which could alternatively be an interface).

The Selection class may feature a coverage property with a value of 'all' or 'selected'.

If the coverage property is set to 'selected', it can also include an items value, represented as an array.

I am looking for a way to prevent the existence of the items property if the coverage is set to 'all'.

Here are some code examples:

const allSelected: Selected = {
  coverage: 'all',
};

allSelected.items = []; // I want this action to trigger an error since it is trying to access `items`.

const onlySelected: Selected = {
  converage: 'selected', // I want this action to trigger an error since it is missing the `items` property.
}

Can the Selection class be defined in this manner?

typescript

Answer 1

Answer №1

If you want to achieve this, you can utilize a discriminated union type:

type Chosen = {
    selection: "chosen",
    options: string[]
} | {
    selection: "all"
};

Your sample code will result in errors as expected.

To access options, you must ensure that selection === "chosen", like so:

function validate(choice: Chosen) {
   // choice.options; // ERROR, options property might not exist
   if(choice.selection === "chosen") {
       console.log(choice.selection); // No error
   }
}

It is not possible to achieve this using the interface keyword, but types and interfaces are essentially interchangeable.

Answer 2

If you want to achieve this, you can utilize a discriminated union type:

type Chosen = {
    selection: "chosen",
    options: string[]
} | {
    selection: "all"
};

Your sample code will result in errors as expected.

To access options, you must ensure that selection === "chosen", like so:

function validate(choice: Chosen) {
   // choice.options; // ERROR, options property might not exist
   if(choice.selection === "chosen") {
       console.log(choice.selection); // No error
   }
}

It is not possible to achieve this using the interface keyword, but types and interfaces are essentially interchangeable.

Is it possible for me to create separate class/interface based on property values?

Here are some code examples:

Answer №1

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