Conditional property for typescript type/object

Question

Let's say I have a recipe ingredient type that looks like this

export type RecipeIngredient = {
  name: string;
  quantity: Number | string;
  measurement: "grams" | "mililitres" | "custom";
};

Example

const potatoes: RecipeIngredient = { name: 'potatoes', quantity: 100, measurement: 'grams' }

which is fine, but I also want to have the quantity as a string only when the measurement is custom. Is this possible?

Correct example

const rosemary: RecipeIngredient = { name: 'rosemary', quantity: 'a handful', measurement: 'custom' }

What I want to be invalid

const fish: RecipeIngredient = { name: 'fish', quantity: '100g', measurement: 'grams' }

hackape · Accepted Answer · 2020-04-09 11:29:43Z

2

export type RecipeIngredient = {
  name: string;
  quantity: Number;
  measurement: "grams" | "mililitres";
} | {
  name: string;
  quantity: string;
  measurement: "custom";
};

answered Apr 9, 2020 at 11:29

hackape

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2 Comments

Jaydeep Over a year ago

Easy but problem is when we want to use that variable, we need to manually cast. for e.g. const custom: RecipeIngredient = { ..., measurement: "custom"}; Now when I use it somewhere, just for example, processIt(custom.measurement as 'custom');

hackape Over a year ago

Well, union type is more useful for annotating function params. For your case I would just let TS infer the type of custom from value instead of annotate it.

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