在以下情况中,是否可以正确地将result
的类型正确推断为boolean
?
interface ActionWithPayload<T extends string, K> { type: T, payload: K }
function ofType<T extends ActionWithPayload<string, any>>(...param: T["type"][]): T extends ActionWithPayload<typeof param[number], infer U> ? U : never {
return null;
}
enum one {
foo = "foo",
bar = "bar"
}
type action = ActionWithPayload<one.foo, boolean> | ActionWithPayload<one.bar, string>;
var result = ofType<action>(one.foo); // type of result should be boolean
答案 0 :(得分:1)
问题在于T["type"]
T
action
时one.foo | one.bar
将function ofType<T extends ActionWithPayload<string, any>, K extends one = one>(...param:K[]): T extends ActionWithPayload<K, infer U> ? U : never {
return null as any;
}
var result = ofType<action, one.foo>(one.foo); // will be boolean
,无论您传递的参数是什么。您需要一个额外的泛型参数,以便编译器推断您传入的枚举成员的文字类型:
one.foo
缺点是您必须显式指定文字类型function ofType<T extends ActionWithPayload<string, any>>() {
return function <K extends one = one>(...param:K[]) : T extends ActionWithPayload<K, infer U> ? U : never{
return null;
}
}
var result = ofType<action>()(one.foo); // will be boolean
,因为您不能只指定其中一个类型参数。作为替代方案,您可以使用双函数方法,因此您可以为第一个函数指定类型参数,并让推理适用于第二个函数:
public class Taxes
{
public int DocumentType { get; set; }
private XElement BuildBodyXML()
{
// other stuff
Address billAddrObj = GetBillTo(dt);
Address buyerAddrObj = GetBuyerPrimary(dt);
var xBillTo = BuildAddress(billAddrObj, "BILL_TO");
var xBuyer = BuildAddress(buyerAddrObj, "BUYER_PRIMARY");
var INVOICE = from row in dt.AsEnumerable()
select new XElement(tcr + "INVOICE",
xBillTo, // This one needs to be conditionally included based on DocumentType
xBuyer,
// ... other elements ...
new XElement(tcr + "INVOICE_NUMBER", row.Field<string>("DOCNUMBR").Trim()));
// other stuff
return INVOICE;
}
public XElement BuildAddress(Address anAddress, string Name)
{
var xAddress = new XElement(tcr + Name);
// other stuff
return xAddress;
}
}