Can the Date class be expanded by overloading the constructor method?

Question

Can the Date class be expanded by overloading the constructor method?

In my dataset, there are dates in different formats that Typescript doesn't recognize. To address this issue, I developed a "safeDateParse" function to handle extended conversions and modified the Date.parse() method accordingly.

/** Custom overload */
Date.parse = function parse(dateAsString: string)  {
    return safeDateParse(dateAsString).valueOf();
};

This modification works well when creating a Date object using the syntax below.

const lastUsedDate = new Date(Date.parse(lastUsed));

However, I also want to override the constructor so I can use the following syntax.

let newDate = new Date(dateString);

I attempted to do this by defining a function, but it didn't produce the desired result as it replaced the Date class instead of extending it. I believe there should be some kind of "super" or "parent" call within it to maintain the core functionality of the Date object.

Is it feasible to override the constructor in this context, and if yes, what would be the correct syntax?

angular typescript

Answer 1

Answer №1

In my opinion, you have the ability to customize the Date constructor in a simple manner:

window.Date = function (original) {
  return function (...params) {
    // Implement any unique operations here
    return new original(...params)
  }
}(Date)

Answer 2

In my opinion, you have the ability to customize the Date constructor in a simple manner:

window.Date = function (original) {
  return function (...params) {
    // Implement any unique operations here
    return new original(...params)
  }
}(Date)

Answer 3

Answer №2

Take a look at this code snippet to get started :

Date.parse = function parse(dateAsString: string) {
    console.log('Custom parse() method executed');
    return safeDateParse(dateAsString).valueOf();
};  

class CustomDate extends Date {
    constructor(dateAsString: string) {
        console.log('Extended Date class created');
        const d = Date.parse(dateAsString);
        super(d);
    }
}

window['Date'] = CustomDate;

new Date();
// Output :
// Extended Date class created
// Custom parse() method executed

Answer 4

Take a look at this code snippet to get started :

Date.parse = function parse(dateAsString: string) {
    console.log('Custom parse() method executed');
    return safeDateParse(dateAsString).valueOf();
};  

class CustomDate extends Date {
    constructor(dateAsString: string) {
        console.log('Extended Date class created');
        const d = Date.parse(dateAsString);
        super(d);
    }
}

window['Date'] = CustomDate;

new Date();
// Output :
// Extended Date class created
// Custom parse() method executed

Answer 5

Answer №3

Just to clarify, I never managed to overload the constructor successfully. After some investigation, I found out that the parameter uses DateConstructor to handle the different overloads. My ideal solution would have been able to handle these overloads. However, in the end, I had to settle for the less preferable solution below and also had to teach my fellow programmers to use Utility.safeDate() instead of new Date()

public static safeDate(dateAsString: string): Date {
    return new Date(safeDateParse(dateAsString));
}

Here is an example call:

const dateObject = Utility.safeDate('șapte mai 1983')

Answer 6

Just to clarify, I never managed to overload the constructor successfully. After some investigation, I found out that the parameter uses DateConstructor to handle the different overloads. My ideal solution would have been able to handle these overloads. However, in the end, I had to settle for the less preferable solution below and also had to teach my fellow programmers to use Utility.safeDate() instead of new Date()

public static safeDate(dateAsString: string): Date {
    return new Date(safeDateParse(dateAsString));
}

Here is an example call:

const dateObject = Utility.safeDate('șapte mai 1983')

Can the Date class be expanded by overloading the constructor method?

Answer №1

Answer №2

Answer №3

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