Different categories combined into a singular category

Question

Different categories combined into a singular category

Trying to define a type that can be one of two options, currently attempting the following:

type TestConfig = {
  file: string;
  name: string;
}

type CakeConfig = {
  run: string;
}

type MixConfig = { test: TestConfig | CakeConfig };

const typeCheck: MixConfig = {
  test: {
    run: 'run!'
  }
};


console.log(typeCheck.test.run)

However, encountering a type checking error:

Property 'run' does not exist on type 'TestConfig | CakeConfig'. Property 'run' does not exist on type 'TestConfig'.

Assumed that using the union operator would allow for either option, but it seems to not work as expected. Any suggestions on how to achieve a similar result? The scenario is where the type could be one of the options without prior knowledge.

Thank you!

typescript

Answer 1

Answer №1

By defining a variable as a union in TypeScript, you allow that variable to hold any value from the specified set of types within the union. TypeScript then ensures that any operation performed on the variable is valid for all possible types it can hold. To handle cases where the type of the configuration may vary at compile-time or dynamically, you can create a helper function that encapsulates the configuration:

type UserData = {
  username: string;
  email: string;
}

type ProductData = {
  id: number;
  price: number;
}

function createData<T extends UserData | ProductData>(data: T): T extends UserData ? UserData : ProductData {
  return data as any; // casting required for generic return
}

createData({ id: 123 }); // ProductData
createData({ username: "user1", email: "user1@example.com" }); // UserData
createData({ price: 50 } as any); // ProductData | UserData

// demonstrating safe inference

const userData = createData({ // UserData
  username: 'john_doe',
  email: 'john.doe@example.com'
});

declare function processUserData(data: UserData | ProductData): void;
processUserData(userData); // valid invocation

declare function processProductData(productData: ProductData): void;
processProductData(userData); // error: argument does not match required type

Answer 2

By defining a variable as a union in TypeScript, you allow that variable to hold any value from the specified set of types within the union. TypeScript then ensures that any operation performed on the variable is valid for all possible types it can hold. To handle cases where the type of the configuration may vary at compile-time or dynamically, you can create a helper function that encapsulates the configuration:

type UserData = {
  username: string;
  email: string;
}

type ProductData = {
  id: number;
  price: number;
}

function createData<T extends UserData | ProductData>(data: T): T extends UserData ? UserData : ProductData {
  return data as any; // casting required for generic return
}

createData({ id: 123 }); // ProductData
createData({ username: "user1", email: "user1@example.com" }); // UserData
createData({ price: 50 } as any); // ProductData | UserData

// demonstrating safe inference

const userData = createData({ // UserData
  username: 'john_doe',
  email: 'john.doe@example.com'
});

declare function processUserData(data: UserData | ProductData): void;
processUserData(userData); // valid invocation

declare function processProductData(productData: ProductData): void;
processProductData(userData); // error: argument does not match required type

Answer 3

Answer №2

When using the typeCheck function, ensure you are doing it correctly:

console.log(typeCheck.test.run)

This code assumes that typeCheck.test has a property called run. However, keep in mind that TestConfig does not have this property. This leads to a contradiction because any variable of type MixConfig can have a test with TestConfig.

To resolve this issue, you need to add the run property to your TestConfig.

Answer 4

When using the typeCheck function, ensure you are doing it correctly:

console.log(typeCheck.test.run)

This code assumes that typeCheck.test has a property called run. However, keep in mind that TestConfig does not have this property. This leads to a contradiction because any variable of type MixConfig can have a test with TestConfig.

To resolve this issue, you need to add the run property to your TestConfig.

Answer 5

Answer №3

To safely access the run property, consider casting typeCheck.test to a CakeConfig like this:

const cakeConfig = typeCheck.test as CakeConfig;
console.log(cakeConfig?.run);

Answer 6

To safely access the run property, consider casting typeCheck.test to a CakeConfig like this:

const cakeConfig = typeCheck.test as CakeConfig;
console.log(cakeConfig?.run);

Answer 7

Answer №4

In agreement with previous responses, the interpreter is unable to guarantee that typeCheck.test is of type CakeConfig.

One potential solution is to create a function that can determine whether or not it is a CakeConfig. This would be particularly beneficial when dealing with external data sources.

function checkIfCakeConfig(config: TestConfig | CakeConfig): config is CakeConfig {
  return config.hasOwnProperty("run")
}

if(checkIfCakeConfig(typeCheck.test))
  console.log(typeCheck.test.run)

Answer 8

In agreement with previous responses, the interpreter is unable to guarantee that typeCheck.test is of type CakeConfig.

One potential solution is to create a function that can determine whether or not it is a CakeConfig. This would be particularly beneficial when dealing with external data sources.

function checkIfCakeConfig(config: TestConfig | CakeConfig): config is CakeConfig {
  return config.hasOwnProperty("run")
}

if(checkIfCakeConfig(typeCheck.test))
  console.log(typeCheck.test.run)

Different categories combined into a singular category

Answer №1

Answer №2

Answer №3

Answer №4

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