TypeScript type for a compare function's parameters
Sebastian Wright
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:
101 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)