Building classes in TypeScript

There is a C# class called Envelope which contains properties for Errors, Paging, and Result. It also has multiple constructors to initialize these properties in different ways.

export class Envelope<T> {

  errors: Error[];
  paging: Paging;
  result: T[];

  constructor(result: T[], paging: Paging, errors: Error[]) {

    this.errors = errors;
    this.paging = paging;
    this.result = result;

  }

}

When trying to convert this class to TypeScript for an Angular 6 project, the issue arises with having multiple constructors as Typescript only allows one. This makes replicating the varied constructors of the original C# class challenging.

One proposed solution is to consider making Envelope an interface in TypeScript instead of a class. This could be a more flexible approach to defining the structure of the Wrapper object used to contain API response elements like Result, Paging, and errors.

Answer №1

To implement constructor overloads, it is necessary to manually distinguish between them in the implementation. In this scenario, the first argument for the implementation would be a combination of T| T[] | Error[]. Type guards can be utilized to differentiate between the cases within the union:

function isErrorArray<T>(e: T | Error[]): e is Error[] {
    return e instanceof Array && e[0] && e[0] instanceof Error;
}
export class Envelope<T> {

    errors: Error[];
    paging: Paging;
    result: T[];

    constructor(result: T)
    constructor(errors: Error[])
    constructor(result: T[], paging: Paging, errors: Error[])
    constructor(result: T | T[] | Error[], paging?: Paging, errors?: Error[]) {
        if (isErrorArray(result)) {
            errors = result;
            result = [] as T[];
        }
        if (Array.isArray(result)) {

        } else {
            result = [result];
        }
        this.errors = errors;
        this.paging = paging;
        this.result = result; // result will be T[] because of the above ifs

    }

}

Answer №2

Your code looks great and could be simplified even further:

export class Envelope<T> {
  constructor(
    public result?: T[], 
    public paging?: Paging, 
    public errors: Error[] = null
  ) {}
}

Here are some shortcuts to keep in mind:

  • If you add an access modifier before a constructor parameter, it automatically becomes a class member without needing to initialize it in the constructor.
  • Adding a ? after a member name makes it optional, resulting in undefined if not provided.
  • By setting a default value in the parameters, you make them optional with a predefined value.

On another note, Typescript supports multiple constructors in a unique way:

export class Envelope<T> {
  constructor(public result: T[])
  constructor(public paging: Paging)
  constructor(public errors: Error[])
  // ...
  {}
}

The downside of this syntax is having to use instanceof to check parameter types, which can be cumbersome compared to using optional parameters.

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