[英]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我正在声明一个 function 用于sort
方法,该方法采用对象数组的值并将其与另一个对象数组进行比较
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:但我在 TypeScript 上这样做,我在 TypeScript 上很新,但我不敢相信我必须这样声明参数:
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?有更简单的方法吗?
Here is the whole code of what I am trying to do:这是我正在尝试做的全部代码:
Thanks for the comment responses.感谢评论回复。
@Chase pointed out there is a shorthand I could use const compare<T extends { role: string }>(a: T, b: T) =>...
@Chase指出有一个简写我可以使用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.但更重要的是@jcalz指出我正在使用sort
方法改变我没有注意到的原始array
。
@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. @jcalz还以更实用的方式重构了代码,并指出如果sort
方法是内联的,则会推断出它们的参数类型。 So this a much better approach.所以这是一个更好的方法。 Not only I don't have to declare the types but is not mutating the original array and it's a lot more readable.我不仅不必声明类型,而且不会改变原始数组,而且它更具可读性。
In the end I used @jcalz approach and did this最后我使用了@jcalz 方法并做了这个
const sortedArray = array
.map(employee => ({
...employee,
id: RoleIndex.find(({ role }) => role.includes(employee.role)) ?. id ?? 100
}))
.sort((a, b) => a.id - b.id)
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