Incorporating a new function into a TypeScript (JavaScript) class method

Question

Incorporating a new function into a TypeScript (JavaScript) class method

Is it possible to add a method to a method within a class?

class MyClass {
    public foo(text: string): string {
        return text + ' FOO!'
    }

    // Looking for a way to dynamically add the method `bar` to `foo`.
}

const obj = new MyClass();
console.log(obj.foo('thing'));
console.log(obj.foo.bar('other thing'));

I came across a similar case involving functions:

function Sum(x: number) { /*...*/ }
namespace Sum {
    export function empty() { /*...*/ }
}

Is there a method to achieve the same functionality with methods of a class without having to patch it after object creation?

javascript typescript methods

Answer 1

Answer №1

To utilize the function within the classes prototype, you have the flexibility to customize it as needed. While it may not be the most visually appealing code, you can implement something along these lines:

class NewClass {
  constructor(myVar) {
    this.myVar = myVar
    this.constructor.prototype.bar.foo = (function(b) {
      console.log(this.myVar, "bar.foo called with:", b)
    }).bind(this)
  }
  bar(message) {
    return message + ' BAR!'
  }

}

const object = new NewClass("John");
object.bar.foo('item')
console.log(object.bar('demo'))

Answer 2

To utilize the function within the classes prototype, you have the flexibility to customize it as needed. While it may not be the most visually appealing code, you can implement something along these lines:

class NewClass {
  constructor(myVar) {
    this.myVar = myVar
    this.constructor.prototype.bar.foo = (function(b) {
      console.log(this.myVar, "bar.foo called with:", b)
    }).bind(this)
  }
  bar(message) {
    return message + ' BAR!'
  }

}

const object = new NewClass("John");
object.bar.foo('item')
console.log(object.bar('demo'))

Answer 3

Answer №2

Although it may not be exactly what you had in mind, the following code accomplishes the desired result:

class MyClass {
  public foo: MyChildClass = new MyChildClass();
}

class MyChildClass {
  public bar(text: string): string{
    return text;
  }
}

const obj = new MyClass();
console.log(obj.foo.bar('thing'));

UPDATE After reading your response to my previous comment, I believe a simpler way to achieve your objective is by using default parameters as shown below:

function foo(text: string, snapshot = false): string{
  if(snapshot){
    return 'snapshot';
  }
  return text;
}

console.debug(foo('test'));
console.debug(foo('test', true));

With this approach, you can clearly see at the point of function call whether you are requesting a bypass or a snapshot due to the additional function names. A similar outcome can be achieved in TypeScript by substituting the arguments of `foo` with an interface containing optional properties. In other programming languages, this technique is often referred to as named parameters:

interface iFooArgs{
  text: string;
  bypass?: boolean;
  snapshot?: boolean;
}

function foo(args: iFooArgs): string {
  if(args.bypass){
    return 'bypass';
  }
  if(args.snapshot){
    return 'snapshot';
  }
  return args.text;
}

console.debug(foo({text: 'test'}));
console.debug(foo({text: 'bypass?', bypass: true}));
console.debug(foo({text: 'snapshot?', snapshot: true}));

Answer 4

Although it may not be exactly what you had in mind, the following code accomplishes the desired result:

class MyClass {
  public foo: MyChildClass = new MyChildClass();
}

class MyChildClass {
  public bar(text: string): string{
    return text;
  }
}

const obj = new MyClass();
console.log(obj.foo.bar('thing'));

UPDATE After reading your response to my previous comment, I believe a simpler way to achieve your objective is by using default parameters as shown below:

function foo(text: string, snapshot = false): string{
  if(snapshot){
    return 'snapshot';
  }
  return text;
}

console.debug(foo('test'));
console.debug(foo('test', true));

With this approach, you can clearly see at the point of function call whether you are requesting a bypass or a snapshot due to the additional function names. A similar outcome can be achieved in TypeScript by substituting the arguments of `foo` with an interface containing optional properties. In other programming languages, this technique is often referred to as named parameters:

interface iFooArgs{
  text: string;
  bypass?: boolean;
  snapshot?: boolean;
}

function foo(args: iFooArgs): string {
  if(args.bypass){
    return 'bypass';
  }
  if(args.snapshot){
    return 'snapshot';
  }
  return args.text;
}

console.debug(foo({text: 'test'}));
console.debug(foo({text: 'bypass?', bypass: true}));
console.debug(foo({text: 'snapshot?', snapshot: true}));

Answer 5

Answer №3

UPDATE: A neat solution to this problem involves the use of intersection types:

class MyClass {

    constructor() {
        this.foo = function () {
            return "this is a"
        };

        this.foo.bar = function() {
            return "this is b"
        };
    }

    public foo: Function & { bar?: () => string };
}

const obj = new MyClass();

console.log(obj.foo());
console.log(obj.foo.bar());

Try it out in the Typescript Playground

This can be achieved by utilizing the & sign in the type declaration of the class method which combines both a Function type and an object type with a single property bar that is also a function. The property being optional in the second type is necessary to satisfy Typescript, even though in practice the assignment within the constructor completes the bar property of foo.

--- PREVIOUS answer below isn't suitable for the intended purpose

Rather than defining a method directly, consider implementing a getter accessor method for the class member 'foo' that returns an object with a function 'bar'

    class MyClass {
        get foo(): {bar: (text: string) => string} {

            // any necessary preliminary code here

            return {
                bar: (text: string) => {
                    return text + ' FOO!';
                }
            }
        }
    }

    const obj = new MyClass();
    console.log(obj.foo.bar('thing'));

Answer 6

UPDATE: A neat solution to this problem involves the use of intersection types:

class MyClass {

    constructor() {
        this.foo = function () {
            return "this is a"
        };

        this.foo.bar = function() {
            return "this is b"
        };
    }

    public foo: Function & { bar?: () => string };
}

const obj = new MyClass();

console.log(obj.foo());
console.log(obj.foo.bar());

Try it out in the Typescript Playground

This can be achieved by utilizing the & sign in the type declaration of the class method which combines both a Function type and an object type with a single property bar that is also a function. The property being optional in the second type is necessary to satisfy Typescript, even though in practice the assignment within the constructor completes the bar property of foo.

--- PREVIOUS answer below isn't suitable for the intended purpose

Rather than defining a method directly, consider implementing a getter accessor method for the class member 'foo' that returns an object with a function 'bar'

    class MyClass {
        get foo(): {bar: (text: string) => string} {

            // any necessary preliminary code here

            return {
                bar: (text: string) => {
                    return text + ' FOO!';
                }
            }
        }
    }

    const obj = new MyClass();
    console.log(obj.foo.bar('thing'));

Incorporating a new function into a TypeScript (JavaScript) class method

Answer №1

Answer №2

Answer №3

Similar questions

Combining plugin setups and functionalities into a unified script

Tips for parsing text responses in React to generate hyperlinks and emphasize specific words

The use of jQuery addClass does not produce any changes

Utilize React and Jest to handle errors by either mocking window values or resolving them

Cancel your subscription to a PubNub channel when the unload event occurs

Using default JavaScriptSerializer to bind DateTime to knockout view model

unable to configure socket.io to start listening on IPv4 address

Why does Axios keep timing out despite successful testing in Postman?

Struggling with ajax: Converting a JavaScript variable to a PHP variable

Need to return to the previous page following submission

How can I use the select2 jQuery plugin with the tags:true option to ensure that selected choices do not appear again in the dropdown menu?

What could be causing this error to occur? I've already got node.js installed on my system

The jQuery UI accordion fails to function properly after refreshing the data

I seem to be having trouble getting the home page to display properly in the app, even though I believe my code is correct. I would greatly appreciate any help in identifying the mistake

How to use JQuery UI sortable to automatically scroll to the bottom of the page

Unable to locate any NativeScript modules for tns-core-module/ui

SyntaxError: Unexpected '<' symbol found in JavaScript file while attempting to import it into an HTML document

Can TSLint and ESLint give a warning when a function is accessed as a property?

Unable to set options, such as the footer template, in Angular UI Typeahead

Error: Attempting to access the 'firstChild' property of a null object when trying to display a street view panorama with Vue.js