I need to sort an array of objects on an optional integer property order in the interval [0, n]. I'm using the JavaScript Array.prototype.sort() for this.

Since order is optional, I know that in my sort I have to confirm that the property exists on the two comparing objects. But since order may equal 0 ( which is falsy), I can't check by the conditional clause (a.order && b.order) - which is sufficient to make TypeScript not complain.

I thought that I could use (a.hasOwnProperty('order') && b.hasOwnProperty('order'), but when I try that conditional clause, TypeScript is complaining that Object is possible undefined. ts(2532).

So somehow hasOwnProperty isn't sufficient to make TypeScript believe that the property actually exists.

At the end of the day, the need is really: How do I sort an array of object by an optional numeric field whose set of possible values may (actually - will) include 0.

But the question at hand is: How do I make TypeScript understand that two objects do have the necessary property for comparison, if I can't use a.property && b.property because either a.property or b.property may be an existing, falsy value (namely, 0).

Edit to add: The condition (a.order >= 0 && b.order >= 0) doesn't work for the same reason.

CodePudding user response：

You have a few options to work with your items in your sorting function:

Use undefined type guard

if (a.order !== undefined && b.order !== undefined) {
}

Use typeof type guard

if (typeof a.order === 'number' && typeof b.order === 'number') {
}

Use a default order value if it does not exist

const aSortOrder = a.order ?? DefaultValue

See Advanced Types > Optional parameters and properties

Note that hasOwnProperty is not implemented to work as a type guard

See Why does Typescript treat `object.hasOwnProperty("key")` as essentially different from `"key" in object`