Creating instance methods in a TypeScript object can be accomplished by defining the methods within the object's class definition. When the object is

Question

Creating instance methods in a TypeScript object can be accomplished by defining the methods within the object's class definition. When the object is

As a seasoned Java developer, I've recently been dabbling in TypeScript.

Let me introduce you to my user object:

export class User
{
    id: string;
    name: string;
    email?: string;
    unit: string;
    street: string;
    postalcode: string;
    paymentmode: string;

    public isComplete(): boolean
    {
        if(this.id != null && this.email != null && this.isAddress()){ return true}
        else return false;
    }

    public isAddress(): boolean
    {
        if(this.street != null && this.postalcode != null){ return true}
        else return false;
    }
}

Now, onto another block of TypeScript...

var user = new User();
user = this.loginService.getLocalUser();

I was initially hoping to call the isComplete method on the user object like this:

user.isComplete()

However, to access it, I realized that I need to treat it as a static object instead:

User.isComplete

Here's how I retrieve my local user object:

getLocalUser(): User {
    var user = new User();
    user = JSON.parse(localStorage.getItem('user'));
    
    return user;
}

Can anyone provide guidance on how to achieve this?

https://i.sstatic.net/enlQx.png

angular typescript

Answer 1

Answer №1

Utilize the following code snippet:

let user = new User().deserialize(this.loginService.getLocalUser());

To enhance your User model, include a method called deserialize():

deserialize(input: any): User {
    let self: any = this;
    for (let prop in input) {
            self[prop] = input[prop];
    }
    return self;
}

When you fetch data from a web service and map it to your User object using parse(), the instance methods are not created because they do not exist within the response object. The purpose of the deserialize() method is to generate a new instance of the User object and transfer values from the response one by one.

Answer 2

Utilize the following code snippet:

let user = new User().deserialize(this.loginService.getLocalUser());

To enhance your User model, include a method called deserialize():

deserialize(input: any): User {
    let self: any = this;
    for (let prop in input) {
            self[prop] = input[prop];
    }
    return self;
}

When you fetch data from a web service and map it to your User object using parse(), the instance methods are not created because they do not exist within the response object. The purpose of the deserialize() method is to generate a new instance of the User object and transfer values from the response one by one.

Answer 3

Answer №2

To start, simply create an instance of the User class using the JSON data obtained from the localStorage.

Update your getLocalUser method with the following code:

getLocalUser(): User {
    let userData = JSON.parse(localStorage.getItem('user'));
    let user = new User(userData);

    return user;
}

Once you have a new instance, you can now access the defined methods of the class as usual:

let user = this.loginService.getLocalUser();
let isComplete = user.isComplete();

console.log(isComplete); // => true/false

Additionally, make sure to include the constructor in the User class:

export class User
{
    id: string;
    name: string;
    // ...

    constructor(userData: any) {
        this.id = userData.id;
        this.name = userData.name;
        // ...
    }

    public isComplete(): boolean
    {
        if(this.id != null && this.email != null && this.isAddress()){ return true}
        else return false;
    }

    // other methods go here...

}

Answer 4

To start, simply create an instance of the User class using the JSON data obtained from the localStorage.

Update your getLocalUser method with the following code:

getLocalUser(): User {
    let userData = JSON.parse(localStorage.getItem('user'));
    let user = new User(userData);

    return user;
}

Once you have a new instance, you can now access the defined methods of the class as usual:

let user = this.loginService.getLocalUser();
let isComplete = user.isComplete();

console.log(isComplete); // => true/false

Additionally, make sure to include the constructor in the User class:

export class User
{
    id: string;
    name: string;
    // ...

    constructor(userData: any) {
        this.id = userData.id;
        this.name = userData.name;
        // ...
    }

    public isComplete(): boolean
    {
        if(this.id != null && this.email != null && this.isAddress()){ return true}
        else return false;
    }

    // other methods go here...

}

Answer 5

Answer №3

The issue arises from this line of code

JSON.parse(localStorage.getItem('user'));

.

The output of this function call does not match the expected type User, therefore it does not have the isComplete member method.

@rattlehand's solution should solve your problem, but it introduces dynamic behavior into the code.

Personally, I believe in having a fully typed solution, as it aligns with one of the key principles of TypeScript.

export interface User
{
    id: string;
    name?: string;
    ...
}

function isComplete(user: User) { 
    if (user.id != null && user.email != null ) { 
        return true;
    }
    return false;
}

function getLocalUser(): User {
    var user = JSON.parse(localStorage.getItem('user')) as User;
    return user;
}

var user = getLocalUser();
console.log(isComplete(user)); // true or false

In TypeScript, using interfaces instead of classes for models may seem unfamiliar to Java developers. However, it is widely adopted in popular frameworks like Angular and RxJs which are written in TS. This practice also extends to DefinitelyTyped's definitions. While there are cases where classes are more suitable, exploring both options can be beneficial.

You can even enhance type safety by creating a generic wrapper for localStorage.

export class LocalStorageService {
    private storage = localStorage;

    get<T>(key: string): T {
        var serialized: string = this.storage.getItem(key);
        return serialized ? <T>JSON.parse(serialized) : undefined;
    }

    set<T>(key: string, data: T) {
        var serialized: string = JSON.stringify(data);
        this.storage.setItem(key, serialized);
    }

    remove(key: string) {
        this.storage.removeItem(key);
    }
}

Answer 6

The issue arises from this line of code

JSON.parse(localStorage.getItem('user'));

.

The output of this function call does not match the expected type User, therefore it does not have the isComplete member method.

@rattlehand's solution should solve your problem, but it introduces dynamic behavior into the code.

Personally, I believe in having a fully typed solution, as it aligns with one of the key principles of TypeScript.

export interface User
{
    id: string;
    name?: string;
    ...
}

function isComplete(user: User) { 
    if (user.id != null && user.email != null ) { 
        return true;
    }
    return false;
}

function getLocalUser(): User {
    var user = JSON.parse(localStorage.getItem('user')) as User;
    return user;
}

var user = getLocalUser();
console.log(isComplete(user)); // true or false

In TypeScript, using interfaces instead of classes for models may seem unfamiliar to Java developers. However, it is widely adopted in popular frameworks like Angular and RxJs which are written in TS. This practice also extends to DefinitelyTyped's definitions. While there are cases where classes are more suitable, exploring both options can be beneficial.

You can even enhance type safety by creating a generic wrapper for localStorage.

export class LocalStorageService {
    private storage = localStorage;

    get<T>(key: string): T {
        var serialized: string = this.storage.getItem(key);
        return serialized ? <T>JSON.parse(serialized) : undefined;
    }

    set<T>(key: string, data: T) {
        var serialized: string = JSON.stringify(data);
        this.storage.setItem(key, serialized);
    }

    remove(key: string) {
        this.storage.removeItem(key);
    }
}

Answer 7

Answer №4

To achieve the desired outcome, follow these steps:

var user = new User;

user.isComplete();

Once you have edited your question and if I understand correctly, you must implement the use of namespace and Triple-Slash Directives in the following manner: Starting with the first file, let's say a.ts:

namespace myNameSpace{
    export class User
    {
        id: string;
        name: string;
        email?: string;
        unit: string;
        street: string;
        postalcode: string;
        paymentmode: string;

        public isComplete(): boolean
        {
            if(this.id != null && this.email != null && this.isAddress()){ return true}
            else return false;
        }

        public isAddress(): boolean
        {
            if(this.street != null && this.postalcode != null){ return true}
            else return false
        }
    }
}

Moving on to the second file, let's call it b.ts:

/// <reference path="a.ts" />
namespace myNameSpace{
    var user = new User();
    //user = this.loginService.getLocalUser(); //this function need to be checked in your environment
    user.isComplete();
}

Answer 8

To achieve the desired outcome, follow these steps:

var user = new User;

user.isComplete();

Once you have edited your question and if I understand correctly, you must implement the use of namespace and Triple-Slash Directives in the following manner: Starting with the first file, let's say a.ts:

namespace myNameSpace{
    export class User
    {
        id: string;
        name: string;
        email?: string;
        unit: string;
        street: string;
        postalcode: string;
        paymentmode: string;

        public isComplete(): boolean
        {
            if(this.id != null && this.email != null && this.isAddress()){ return true}
            else return false;
        }

        public isAddress(): boolean
        {
            if(this.street != null && this.postalcode != null){ return true}
            else return false
        }
    }
}

Moving on to the second file, let's call it b.ts:

/// <reference path="a.ts" />
namespace myNameSpace{
    var user = new User();
    //user = this.loginService.getLocalUser(); //this function need to be checked in your environment
    user.isComplete();
}

Answer 9

Answer №5

Why not simplify it like this:

fetchLocalUser(): User {
    return Object.assign(new User(), JSON.parse(localStorage.getItem('user')));
}

After using MDN: Object.assign(), you essentially:

take the values of all enumerable own properties from one or more source objects and merge them into a target object.

Furthermore,

this method will result in the target object being returned.

This way, you will have a freshly instantiated version of the User with all the properties from the parsed JSON data.

Answer 10

Why not simplify it like this:

fetchLocalUser(): User {
    return Object.assign(new User(), JSON.parse(localStorage.getItem('user')));
}

After using MDN: Object.assign(), you essentially:

take the values of all enumerable own properties from one or more source objects and merge them into a target object.

Furthermore,

this method will result in the target object being returned.

This way, you will have a freshly instantiated version of the User with all the properties from the parsed JSON data.

Answer 11

Answer №6

Give this a shot

let currentUser = (<User>this.loginService.getLocalUser());

Answer 12

Give this a shot

let currentUser = (<User>this.loginService.getLocalUser());

Creating instance methods in a TypeScript object can be accomplished by defining the methods within the object's class definition. When the object is

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

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