使用Typescript，键入guard和reflection来根据类型验证对象

Question

I am trying to DRY, by using ES6, Typescript, I have created an interface, call it IUser which has properties.

• What is the best way to know if an object conforms to the interface at runtime?
• How can I get a list of the errors? The IDE knows which properties are missing, why I can't?

I have seen examples using reflection and other techniques which test one property at a time. But there must be a way to easily know if an object implements all properties, and if not which ones.

IKnowing that some properties are optional '?' and we know the type, this should be possible!

type IUser = {
    name: string
    age?: number
    gender: string | any
}

// Assume this object gets created at runtime.
let john = {
    name: "John"
}

let errors = ValidateThis(john, IUser)
// Return something like this ["Error: Missing value gender", "Warning: Missing optional value age"]

• I don't want to specify each property, so there must be a way to iterate the type properties.
• I don't want to use any of the existing npm projects I found which force me to create, yet a separate schema file or use a class.

Answer 1

You aren't the first person to ask for this. Interfaces don't exist at runtime, because TS fully erases all type information during compilation.

However, ECMAScript has standards-tracked a reflection API, and TypeScript is on-board with it. Your best bet for accomplishing this is to use experimentalDecoratorMetadata to automatically decorate your classes with type information, and then use the reflection API at runtime to validate. (Ctrl-f and search for experimentalDecoratorMetadata ).

Answer 2

This doesn't exist. You should use one of the existing NPM projects, or write yet another.