Persuade TypeScript to trust that all necessary keys will be present in an object

Question

Persuade TypeScript to trust that all necessary keys will be present in an object

I find myself in this particular scenario:

const user: UserObj = User.get(userId);

if ([user.foo, user.bar, user.baz].some((k) => !k))
  throw new Error(`Missing fields for user ${userId}`);

const createParams: CreateParams = {
  firstName: user.firstName,
  lastName: user.lastName,
  email: user.email,
  phone: user.phone,
  foo: user.foo,
  bar: user.bar,
  baz: user.baz
};

My issue here is that foo bar baz need to be present in CreateParams. TypeScript is raising errors because those keys are optional in UserObj. However, I have implemented a .some check to ensure they exist, but TypeScript still shows errors. How can I prove to TypeScript that these required keys will always be provided?

typescript

Answer 1

Answer №1

If you want to handle this situation differently, there are a few approaches you can take. One option is to move the guard logic into a separate function:

interface HasOptional extends UserObj {
    foo: string,
    bar: string,
    baz: string,
}

function hasOptional(user: UserObj): user is HasOptional {
    return [user.foo, user.bar, user.baz].every(k => k);
}

if (!hasOptional(user))
    throw new Error('missing fields');

Alternatively, you can adjust the conditional statement to make it clearer for TypeScript to interpret:

if (!user.foo || !user.bar || !user.baz)
    throw new Error('missing fields');

It's worth noting that relying on "truthiness" can sometimes lead to unexpected results, as !"" and !0 evaluate to true. Using .some(k => !k) to check for missing properties will also capture empty strings. Depending on your requirements, you may want to consider one of the following alternatives:

// Suitable if 0/''/false are valid values for foo/bar/baz.
function hasOptional(user: UserObj): user is HasOptional {
    return [user.foo, user.bar, user.baz].every(k != null);
}

// Also effective if null is a valid value for foo/bar/baz.
function hasOptional(user: UserObj): user is HasOptional {
    return [user.foo, user.bar, user.baz].every(k !== undefined);
}

// Works even if undefined is a valid value for foo/bar/baz.
function hasOptional(user: UserObj): user is HasOptional {
    return ['foo', 'bar', 'baz'].every(k in user);
}

Answer 2

If you want to handle this situation differently, there are a few approaches you can take. One option is to move the guard logic into a separate function:

interface HasOptional extends UserObj {
    foo: string,
    bar: string,
    baz: string,
}

function hasOptional(user: UserObj): user is HasOptional {
    return [user.foo, user.bar, user.baz].every(k => k);
}

if (!hasOptional(user))
    throw new Error('missing fields');

Alternatively, you can adjust the conditional statement to make it clearer for TypeScript to interpret:

if (!user.foo || !user.bar || !user.baz)
    throw new Error('missing fields');

It's worth noting that relying on "truthiness" can sometimes lead to unexpected results, as !"" and !0 evaluate to true. Using .some(k => !k) to check for missing properties will also capture empty strings. Depending on your requirements, you may want to consider one of the following alternatives:

// Suitable if 0/''/false are valid values for foo/bar/baz.
function hasOptional(user: UserObj): user is HasOptional {
    return [user.foo, user.bar, user.baz].every(k != null);
}

// Also effective if null is a valid value for foo/bar/baz.
function hasOptional(user: UserObj): user is HasOptional {
    return [user.foo, user.bar, user.baz].every(k !== undefined);
}

// Works even if undefined is a valid value for foo/bar/baz.
function hasOptional(user: UserObj): user is HasOptional {
    return ['foo', 'bar', 'baz'].every(k in user);
}

Answer 3

Answer №2

If TypeScript is unsure about your validation, there may not be another solution. You can use @ts-ignore or ||.

//@ts-ignore
foo: user.foo
// or
bar: user.bar || ''

If you're not happy with that, you can create a new type to confirm that typechecking has been performed.

For instance:

// interface TypeChecked extends User
type TypeChecked {
   foo: string;
   bar: boolean;
   baz: any;
}

const createParams: CreateParams = {
  firstName: user.firstName,
  lastName: user.lastName,
  email: user.email,
  phone: user.phone,
  foo: (<TypeChecked> user).foo,
  bar: (<TypeChecked> user).bar,
  baz: (<TypeChecked> user).baz
};

Answer 4

If TypeScript is unsure about your validation, there may not be another solution. You can use @ts-ignore or ||.

//@ts-ignore
foo: user.foo
// or
bar: user.bar || ''

If you're not happy with that, you can create a new type to confirm that typechecking has been performed.

For instance:

// interface TypeChecked extends User
type TypeChecked {
   foo: string;
   bar: boolean;
   baz: any;
}

const createParams: CreateParams = {
  firstName: user.firstName,
  lastName: user.lastName,
  email: user.email,
  phone: user.phone,
  foo: (<TypeChecked> user).foo,
  bar: (<TypeChecked> user).bar,
  baz: (<TypeChecked> user).baz
};

Persuade TypeScript to trust that all necessary keys will be present in an object

Answer №1

Answer №2

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