Closed hemedani closed 1 month ago
This task was not very difficult. By writing a few helper functions, I was able to get the output I wanted. However, there is a serious problem. In Runtime, I did not find anything related to a field being optional. Actually, I have written a framework called Lesan with Superstruct
. You can see the code I wrote to create the types here:
https://github.com/MiaadTeam/lesan/blob/main/src/types/generateTypesFromStruct.ts
Of course, this code has not been fully written yet.
But the problem is when we create an object with the following two fields:
object({
optionalName: optional(string()),
name: string(),
});
At runtime, we get something like this:
{
"type": "object",
"schema": {
"optionalName": {
"type": "string",
"schema": null,
},
"name": {
"type": "string",
"schema": null,
},
},
}
I don't see anything related to the optionalName
field being optional! While, we have put it in the optional(string())
function.
How should I know at runtime if a field is optional?
Finally, the type I create for the above struct is as follows:
{
optionalName: string;
name: string;
}
which I want to create it like this:
{
optionalName?: string;
name: string;
}
typeof user.TYPE
Completed, but how? Please give a brief explanation. thanks a lot
import { number, object, optional, string } from "superstruct";
const user = object({
id: number(),
name: string(),
some: optional(string()),
});
console.log(user.schema.name.validate(undefined)[0] == null) // false
console.log(user.schema.some.validate(undefined)[0] == null) // true
I want to create
.d.ts
file from superstruct for example I want to convert:to
And write it to a file with Node.js or Deno, but
Infer<typeof user>
does not available in runtime.