Typescript Interface - Possible to make "one or the other" properties required?

You can use a union type to do this:

interface MessageBasics {
  timestamp?: number;
  /* more general properties here */
}
interface MessageWithText extends MessageBasics {
  text: string;
}
interface MessageWithAttachment extends MessageBasics {
  attachment: Attachment;
}
type Message = MessageWithText | MessageWithAttachment;

If you want to allow both text and attachment, you would write

type Message = MessageWithText | MessageWithAttachment | (MessageWithText & MessageWithAttachment);

If you're truly after "one property or the other" and not both you can use never in the extending type:

interface MessageBasics {
  timestamp?: number;
  /* more general properties here */
}
interface MessageWithText extends MessageBasics {
  text: string;
  attachment?: never;
}
interface MessageWithAttachment extends MessageBasics {
  text?: never;
  attachment: string;
}
type Message = MessageWithText | MessageWithAttachment;

// 👍 OK 
let foo: Message = {attachment: 'a'}

// 👍 OK
let bar: Message = {text: 'b'}

// ❌ ERROR: Type '{ attachment: string; text: string; }' is not assignable to type 'Message'.
let baz: Message = {attachment: 'a', text: 'b'}

Example in Playground

You can go deeper with @robstarbuck solution creating the following types:

type Only<T, U> = {
  [P in keyof T]: T[P];
} & {
  [P in keyof U]?: never;
};

type Either<T, U> = Only<T, U> | Only<U, T>;

And then Message type would look like this

interface MessageBasics {
  timestamp?: number;
  /* more general properties here */
}
interface MessageWithText extends MessageBasics {
  text: string;
}
interface MessageWithAttachment extends MessageBasics {
  attachment: string;
}
type Message = Either<MessageWithText, MessageWithAttachment>;

With this solution you can easily add more fields in MessageWithText or MessageWithAttachment types without excluding it in another.

Thanks @ryan-cavanaugh that put me in the right direction.

I have a similar case, but then with array types. Struggled a bit with the syntax, so I put it here for later reference:

interface BaseRule {
  optionalProp?: number
}

interface RuleA extends BaseRule {
  requiredPropA: string
}

interface RuleB extends BaseRule {
  requiredPropB: string
}

type SpecialRules = Array<RuleA | RuleB>

// or

type SpecialRules = (RuleA | RuleB)[]

// or (in the strict linted project I'm in):

type SpecialRule = RuleA | RuleB
type SpecialRules = SpecialRule[]

Update:

Note that later on, you might still get warnings as you use the declared variable in your code. You can then use the (variable as type) syntax. Example:

const myRules: SpecialRules = [
  {
    optionalProp: 123,
    requiredPropA: 'This object is of type RuleA'
  },
  {
    requiredPropB: 'This object is of type RuleB'
  }
]

myRules.map((rule) => {
  if ((rule as RuleA).requiredPropA) {
    // do stuff
  } else {
    // do other stuff
  }
})