Accessing different properties in a typescript uni

问题:

I'm creating a function that handles objects from the database. I have two different data structures where the same property has a different name. I can't change that, so I have to handle it in JavaScript.
The objects have other differences, but that's not important to this function.
I want to use the same function for two different types of objects. Here's sample code demonstrating my problem:

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;
}

Of course, this code runs. But the IDE marks both the lines name = item.itemName; and name = item.itemTitle; as errors ("Property does not exist on type"), because both types do not have both properties.

So, what's the proper typescript way to do this?

回答1:

You need to create a User Defined Type Guard, then you can use an if statement and get the correct typing.

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

Then you can get the typing much cleaner:

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

Check it out here. Item is correctly cast to either TypeA or TypeB.

回答2:

Intellij accepts this syntax:

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

the official way according to the typescript docs is this: https://www.typescriptlang.org/docs/handbook/functions.html

回答3:

you can make a type assertion if you don't do this too often :

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

or

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

if you need to make this check more than once or twice, you'd better writing a type guard as @Daryl suggests.

回答4:

Use typeguards:

interface TypeA {
    itemName: string;
}

interface TypeB {
    itemTitle: string;
}

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

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

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

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

    return name;
}

回答5:

I won't complicate things. If you're really sure that your object has either the one or the other property, a name = item['itemName'] || item['itemTitle'] or name = item.hasOwnProperty('itemName') ? item['itemName'] : item['itemTitle'] would be sufficient.

Note that TypeScript usually stops complaining if you access properties using the bracket notation instead of the dot notation. I would suggest adding a comment, though.