Loop through a collection of elements of a certain kind and selectively transfer only certain items to a different collection of a different kind

Question

Loop through a collection of elements of a certain kind and selectively transfer only certain items to a different collection of a different kind

When working with typescript, I am faced with the challenge of dealing with two arrays:

interface IFirst{
    name: string;
    age: number
}

interface ISecond {
    nickName: string;
    lastName: string;
}   

myFirstArray: IFirst[];
mySecondArray: ISecond[];

myFirstArray = [
    {name: 'tim', age: 20},
    {name: 'jim', age: 30}
]

I am trying to figure out a way to loop through myFirstArray and assign all names as nicknames in mySecondArray. Is there a method to achieve this functionality?

this.myFirstArray.forEach((element: IFirst) => {
        this.mySecondArray.push({nickName = element.name});
      })

arrays angular typescript

Answer 1

Answer №1

If you only want to set a nickname, you can make the lastName field optional in the ISecond interface by using lastName?: string. The updated interface will look like this:

interface ISecond {
  nickName: string;
  lastName?: string;
}

To assign the nickname value from each element of myFirstArray to mySecondArray, you can utilize the this.myFirstArray.map() method. Here's how you can do it:

this.mySecondArray = this.myFirstArray.map((element: IFirst) => {
  return { nickName: element.name };
});

console.log("this.mySecondArray ", this.mySecondArray);

Check out the working code on Stackblitz... (View console for output).

Answer 2

If you only want to set a nickname, you can make the lastName field optional in the ISecond interface by using lastName?: string. The updated interface will look like this:

interface ISecond {
  nickName: string;
  lastName?: string;
}

To assign the nickname value from each element of myFirstArray to mySecondArray, you can utilize the this.myFirstArray.map() method. Here's how you can do it:

this.mySecondArray = this.myFirstArray.map((element: IFirst) => {
  return { nickName: element.name };
});

console.log("this.mySecondArray ", this.mySecondArray);

Check out the working code on Stackblitz... (View console for output).

Answer 3

Answer №2

In this scenario, mySecondArray is defined as an array of type ISecond. Therefore, when assigning values to it, both properties of the object must be included.

For example -

this.myFirstArray.map((element: IFirst) => {
  this.mySecondArray.push({nickName = element.name, lastName: ''});
})

Answer 4

In this scenario, mySecondArray is defined as an array of type ISecond. Therefore, when assigning values to it, both properties of the object must be included.

For example -

this.myFirstArray.map((element: IFirst) => {
  this.mySecondArray.push({nickName = element.name, lastName: ''});
})

Answer 5

Answer №3

Utilize the map function to transform your array:

this.peopleList = [
    {name: 'Alice', age: 25},
    {name: 'Bob', age: 35}
];
this.modifiedPeopleList = this.peopleList.map(({name, age})=> <IModified><any>{ 
    fullName: name,
    yearsOld: age
    });
console.log(this.modifiedPeopleList);

Answer 6

Utilize the map function to transform your array:

this.peopleList = [
    {name: 'Alice', age: 25},
    {name: 'Bob', age: 35}
];
this.modifiedPeopleList = this.peopleList.map(({name, age})=> <IModified><any>{ 
    fullName: name,
    yearsOld: age
    });
console.log(this.modifiedPeopleList);

Answer 7

Answer №4

Make sure to properly set up your second array:

friends: IFriend[] = [];

Once initialized, populate it by iterating through the first array:

this.users.forEach((user: IUser) => {
    this.friends.push({username: user.name, lastName: ''});
})

Answer 8

Make sure to properly set up your second array:

friends: IFriend[] = [];

Once initialized, populate it by iterating through the first array:

this.users.forEach((user: IUser) => {
    this.friends.push({username: user.name, lastName: ''});
})

Loop through a collection of elements of a certain kind and selectively transfer only certain items to a different collection of a different kind

Answer №1

Answer №2

Answer №3

Answer №4

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