Exploring various attributes within a TypeScript union declaration

Question

Exploring various attributes within a TypeScript union declaration

I've encountered a challenge while creating a function that interacts with database objects. In this case, I have two different data structures where the same property is named differently. Unfortunately, I cannot change these names and must find a solution using JavaScript.
Although there are additional variations between the objects, they are not relevant to the specific function at hand.
My goal is to utilize the same function for handling both types of objects. Below is a snippet demonstrating the issue:

interface TypeA {
    itemName: string;
}

interface TypeB {
    itemTitle: string;
}

function getItemName(item: TypeA | TypeB): string {
    let name = '';

    if (item.hasOwnProperty('itemName')) {
        name = item.itemName;
    } else {
        name = item.itemTitle;
    }

    return name;
}

While the code functions as intended, my IDE highlights errors on the lines name = item.itemName; and name = item.itemTitle;, stating "Property does not exist on type" due to the lack of both properties in each object type.

What would be the correct approach in TypeScript to address this dilemma?

if-statement typescript object-properties

Answer 1

Answer №1

To ensure correct typing, it is essential to create a User Defined Type Guard and utilize an if statement.

function isTypeA(value: TypeA | TypeB): value is TypeA {
    return value.hasOwnProperty('itemName');
}

This approach allows for cleaner typing:

function getItemName(item: TypeA | TypeB): string {
    return isTypeA(item) ? item.itemName : item.itemTitle;
}

Take a look here. The item is appropriately cast as either TypeA or TypeB.

Answer 2

To ensure correct typing, it is essential to create a User Defined Type Guard and utilize an if statement.

function isTypeA(value: TypeA | TypeB): value is TypeA {
    return value.hasOwnProperty('itemName');
}

This approach allows for cleaner typing:

function getItemName(item: TypeA | TypeB): string {
    return isTypeA(item) ? item.itemName : item.itemTitle;
}

Take a look here. The item is appropriately cast as either TypeA or TypeB.

Answer 3

Answer №2

While I may not be right on time, consider implementing the following code snippet in your function:

if ('productName' in product) {
    title = product.productName;
} else {
    title = product.productTitle;
}

Answer 4

While I may not be right on time, consider implementing the following code snippet in your function:

if ('productName' in product) {
    title = product.productName;
} else {
    title = product.productTitle;
}

Answer 5

Answer №3

When you find yourself in a situation where you need to perform a type assertion, it's important to do so judiciously:

if (item.hasOwnProperty('itemName')) {
    name = (item as TypeA).itemName;
} else {
    name = (item as TypeB).itemTitle;
}

Alternatively, you can also use the following syntax:

if (item.hasOwnProperty('itemName')) {
    name = (<TypeA>item).itemName;
} else {
    name = (<TypeB>item).itemTitle;
}

If you anticipate needing to perform this check frequently, it may be more beneficial to create a type guard as recommended by @Daryl.

Answer 6

When you find yourself in a situation where you need to perform a type assertion, it's important to do so judiciously:

if (item.hasOwnProperty('itemName')) {
    name = (item as TypeA).itemName;
} else {
    name = (item as TypeB).itemTitle;
}

Alternatively, you can also use the following syntax:

if (item.hasOwnProperty('itemName')) {
    name = (<TypeA>item).itemName;
} else {
    name = (<TypeB>item).itemTitle;
}

If you anticipate needing to perform this check frequently, it may be more beneficial to create a type guard as recommended by @Daryl.

Answer 7

Answer №4

interface TypeA {
  a: string
}
interface TypeB {
  b: string
}

const demoFunction = (input: TypeA | TypeB): string => {
  return (input as TypeA).a || (input as TypeB).b;
}

demoFunction({ a: 'Greetings' }); // 'Greetings'
demoFunction({ b: 'Universe' }); // 'Universe'

Answer 8

interface TypeA {
  a: string
}
interface TypeB {
  b: string
}

const demoFunction = (input: TypeA | TypeB): string => {
  return (input as TypeA).a || (input as TypeB).b;
}

demoFunction({ a: 'Greetings' }); // 'Greetings'
demoFunction({ b: 'Universe' }); // 'Universe'

Answer 9

Answer №5

When using Intellij, the following syntax is accepted:

function getItemName(item: TypeA): string;
function getItemName(item: TypeB): string;
function getItemName(item): string {
    return item.hasOwnProperty('itemName') ? item.itemName : item.itemTitle;
}

According to the TypeScript documentation, the official way is outlined here: https://www.typescriptlang.org/docs/handbook/functions.html

Answer 10

When using Intellij, the following syntax is accepted:

function getItemName(item: TypeA): string;
function getItemName(item: TypeB): string;
function getItemName(item): string {
    return item.hasOwnProperty('itemName') ? item.itemName : item.itemTitle;
}

According to the TypeScript documentation, the official way is outlined here: https://www.typescriptlang.org/docs/handbook/functions.html

Answer 11

Answer №6

I'll keep it simple. If you are absolutely certain that your object has either one property or the other, you can use either

name = item['itemName'] || item['itemTitle']

or

name = item.hasOwnProperty('itemName') ? item['itemName'] : item['itemTitle']

, and that should do the trick.

It's worth mentioning that TypeScript tends to be less picky if you access properties using bracket notation instead of dot notation. Still, adding a comment for clarity wouldn't hurt.

Answer 12

I'll keep it simple. If you are absolutely certain that your object has either one property or the other, you can use either

name = item['itemName'] || item['itemTitle']

or

name = item.hasOwnProperty('itemName') ? item['itemName'] : item['itemTitle']

, and that should do the trick.

It's worth mentioning that TypeScript tends to be less picky if you access properties using bracket notation instead of dot notation. Still, adding a comment for clarity wouldn't hurt.

Answer 13

Answer №7

Utilize typeguards in your TypeScript code:

interface TypeA {
    itemName: string;
}

interface TypeB {
    itemTitle: string;
}

function isTypeA(value: any): value is TypeA
{
    return value.hasOwnProperty('itemName');
}

function isTypeB(value: any): value is TypeB
{
    return value.hasOwnProperty('itemTitle');
}

function getItemName(item: TypeA | TypeB): string
{
    let name = '';

    if (isTypeA(item))
    {
        name = item.itemName;
    }
    else
    {
        name = item.itemTitle;
    }

    return name;
}

Answer 14

Utilize typeguards in your TypeScript code:

interface TypeA {
    itemName: string;
}

interface TypeB {
    itemTitle: string;
}

function isTypeA(value: any): value is TypeA
{
    return value.hasOwnProperty('itemName');
}

function isTypeB(value: any): value is TypeB
{
    return value.hasOwnProperty('itemTitle');
}

function getItemName(item: TypeA | TypeB): string
{
    let name = '';

    if (isTypeA(item))
    {
        name = item.itemName;
    }
    else
    {
        name = item.itemTitle;
    }

    return name;
}

Exploring various attributes within a TypeScript union declaration

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

Answer №7

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