问题
How do I say that I want an interface to be one or the other, but not both or neither?
interface IFoo {
bar: string /*^XOR^*/ can: number;
}
回答1:
As proposed in this issue, you can use conditional types (introduced in Typescript 2.8) to write a XOR type:
type Without<T, U> = { [P in Exclude<keyof T, keyof U>]?: never };
type XOR<T, U> = (T | U) extends object ? (Without<T, U> & U) | (Without<U, T> & T) : T | U;
And you can use it like so:
type IFoo = XOR<{bar: string;}, {can: number}>;
let test: IFoo;
test = { bar: "test" } // OK
test = { can: 1 } // OK
test = { bar: "test", can: 1 } // Error
test = {} // Error
回答2:
You can use union types along with the never type to achieve this:
type IFoo = {
bar: string; can?: never
} | {
bar?: never; can: number
};
let val0: IFoo = { bar: "hello" } // OK only bar
let val1: IFoo = { can: 22 } // OK only can
let val2: IFoo = { bar: "hello", can: 22 } // Error foo and can
let val3: IFoo = { } // Error neither foo or can
回答3:
You can get "one but not the other" with union and optional void type:
type IFoo = {bar: string; can?: void} | {bar?:void; can: number};
However, you have to use --strictNullChecks to prevent having neither.
回答4:
try this:
type Foo = {
bar?: void;
foo: string;
}
type Bar = {
foo?: void;
bar: number;
}
type FooBar = Foo | Bar;
// Error: Type 'string' is not assignable to type 'void'
let foobar: FooBar = {
foo: "1",
bar: 1
}
// no errors
let foo = {
foo: "1"
}
来源：https://stackoverflow.com/questions/44425344/typescript-interface-with-xor-barstring-xor-cannumber