Conditional types in TypeScript

Question

I was wondering if I can have conditional types in TypeScript?

Currently I have the following interface:

interface ValidationResult {   isValid: boolean;   errorText?: string; } 

But I want to remove errorText, and only have it when isValid is false as a required property.

I wish I was able to write it as the following interface:

interface ValidationResult {   isValid: true; }  interface ValidationResult {   isValid: false;   errorText: string; } 

But as you know, it is not possible. So, what is your idea about this situation?

Answer 1

One way to model this kind of logic is to use a union type, something like this

interface Valid {   isValid: true }  interface Invalid {   isValid: false   errorText: string }  type ValidationResult = Valid | Invalid  const validate = (n: number): ValidationResult => {   return n === 4 ? { isValid: true } : { isValid: false, errorText: "num is not 4" } } 

The compiler is then able to narrow the type down based on the boolean flag

const getErrorTextIfPresent = (r: ValidationResult): string | null => {   return r.isValid ? null : r.errorText }