Typescript Type for a Compare Function's Parameters
I Am Declaring a Function to Use on a Sort Method That Takes the Value on an Array of Objects and Compares It Against Another Array of Objects Const Compare =...
I am declaring a function to use on a sort method that takes the value on an array of objects and compares it against another array of objects
const compare = (a, b) => {
let x = 0
let y = 0
RoleIndex.forEach(({ role, id}) => {
if (role.includes(a.role)) x = id
if (role.includes(b.role)) y = id
})
if (x > y) return 1
else return -1
})
But I am doing this on TypeScript, I am quite new at TypeScript but I can not believe I have to declare the parameters this way:
const compare = ((a: { role: string}, b: { role: string}) => {
let x: number = 0
let y: number = 0
RoleIndex.forEach(({ role, id}) => {
if (role.includes(a.role)) x = id
if (role.includes(b.role)) y = id
})
if (x > y) return 1
else return -1
})
Is there a simpler way?
Here is the whole code of what I am trying to do:
1 Answer
Thanks for the comment responses.
@Chase pointed out there is a shorthand I could use const compare<T extends { role: string }>(a: T, b: T) =>...
But more importantly @jcalz pointed out I was mutating the original array which I did not notice, with the sort method.
@jcalz also refactored the code in a much more functional way, and pointed out that if the sort method is inline their parameters types are inferred. So this a much better approach. Not only I don't have to declare the types but is not mutating the original array and it's a lot more readable.
In the end I used @jcalz approach and did this
const sortedArray = array
.map(employee => ({
...employee,
id: RoleIndex.find(({ role }) => role.includes(employee.role)) ?. id ?? 100
}))
.sort((a, b) => a.id - b.id)
David Miller brings 15 years of experience in global economics, personal finance strategy, and market dynamics. He specializes in turning complex economic trends into actionable insights for everyday readers.
