从通用参数类型确定 typescript 属性联合

Question

我有一些过滤操作

type FilterOp = 'Equals' | 'NotEquals' | 'Greater' | 'GreaterEqual' | 'Less' | 'LessEqual';
type ArrayFilterOp = 'In' | 'NotIn';
type StringFilterOp = 'StartsWith' | 'EndsWith' | 'Contains' | 'NotContains';
type DateFilterOp = 'DateIn' | 'DateNotIn'

这是我当前的过滤器定义

type GenericFilter<T> = {
    Property: string,
    Value: T,
    Operation: FilterOp | ArrayFilterOp | StringFilterOp | DateFilterOp
}

是否可以根据T参数以某种方式确定Operation类型？ 例如，如果我的T是Date那么 typescript 将只允许将DateFilterOp和FilterOp分配给Operation属性

let dateFilter: GenericFilter<Date> = {
    Property: "DateCreated",
    Value: new Date(),
    Operation: // now I can only set value from FilterOp or DateFilterOp
}

Answer 1

假设我正确理解了约束，您可以使用条件类型来表示它：

type Operation<T> =
  | FilterOp
  | (T extends Array<any> ? ArrayFilterOp : never)
  | (T extends string ? StringFilterOp : never)
  | (T extends Date ? DateFilterOp : never);

type GenericFilter<T> = {
  Property: string;
  Value: T;
  Operation: Operation<T>;
};

这将支持您声明的用例：

let dateFilter: GenericFilter<Date> = {
  Property: "DateCreated",
  Value: new Date(),
  Operation: "DateIn"
};

以及与帮助器 function 一起工作，这样您就可以推断T而不是手动指定它：

const asGenericFilter = <T>(filt: GenericFilter<T>) => filt;

let stringFilter = asGenericFilter({
  Property: "Name",
  Value: "Alice",
  Operation: "StartsWith" // hinted as Operation<string>
})

希望有帮助； 祝你好运！

链接到代码

Answer 2

我已经设法通过OperationMap实施解决方案

type OperationMap<T> = 
    T extends Date
        ? (DateFilterOp | FilterOp)
        : T extends Array<any>
            ? ArrayFilterOp
            : T extends String
                ? FilterOp | StringFilterOp
    : FilterOp;

type GenericFilter<T> = {
    Property: string,
    Value: T,
    Operation: OperationMap<T>
}