How to console.log typescript type

Question

I know that typescript type is not runtime variable. But is there any workaround how to print type name and variables?

type TestDto = {
    active: number;  
    user_name: string;
};

console.log(TestDto);

throw only refers to a type, but is being used as a value here.

I would like to create object property based on TS type. I would like to create JS object with property active and user_name with some default values

Answer 1

I would like to create object property based on TS type. I would like to create JS object with property active and user_name with some default values.

Your better option than creating an object based on a type to do the opposite and create the type based on an object. This not only gives you something concrete you can use at runtime, it can also serve as the defaults for the properties:

const DefaultTestDto = {
    active: -1,
    user_name: "",
}

type TestDto = typeof DefaultTestDto;


const newDto1: TestDto = {...DefaultTestDto};
newDto1.user_name = "Fred";
//or
const newDto2: TestDto = Object.assign({}, DefaultTestDto, {user_name: "Barney"});

console.log(newDto1); // { "active": -1, "user_name": "Fred" } 
console.log(newDto2); // { "active": -1, "user_name": "Barney" }

Playground Link

Answer 2

As you wrote, types are removed before execution. If you want to create an object based on a type, classes are fine for that. Or most IDE tools show you the structure of types when you hover over them if that's your purpose.