CODEWITHSUNDEEP
javascripttypescript
Jon Sud
Jon Sud

Reputation: 11641

Can I add property to object only if true in typescript?

I have an object:

const options = {};

and I want to add fields base on another object. if they false I don't want to add this key to the object.

right now I write my code like this:

const options = {};
if (foo.nullable) options.nullable = true;
if (foo.baz) options.baz = true;
if (foo.bar) options.bar = true;
...

if I write like this I get the nullable key anyway. no matter if is true or false and I want only if true the key should exist.

 const options = { nullable: options.nullable }
 // options = { nullable: ... }

So I asking is there a way to write this in better in typescript? something like:

 const options = { nullable && options.nullable } // should be options = {} if nullable is false. otherwise is options = { nullable: true }

Upvotes: 1

Views: 1723

Answers (1)

T.J. Crowder
T.J. Crowder

Reputation: 1074585

So I asking is there a way to write this in better in typescript?

Not really, no. What you have with the if statements is probably simplest.

You could use spread notation, like this:

const options = {
    ...(foo.nullable ? {nullable: true} : undefined),
    ...(foo.baz ? {baz: true} : undefined),
    ...(foo.bar ? {bar: true} : undefined),
};

That works because if you spread undefined it doesn't do anything. But I don't think I'd call it "better." :-)

If the values of your foo fields are always primitives, you could use && but I wouldn't:

const options = {
    ...foo.nullable && {nullable: true},
    ...foo.bar && {bar: true},
    ...foo.baz && {baz: true},
};

But if the foo fields might be objects, that's not going to work, and clarity probably suffers anyway. (That "works" for primitives because if you spread a primitive, it gets converted to the equivalent object — 42 becomes new Number(42), in effect — and by default the object types for primitives have no own, enumerable properties, so nothing gets spread.)

Upvotes: 1

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