Ways to exclude an object field in Angular 7

Question

Ways to exclude an object field in Angular 7

Is there a way in Angular to exclude specific fields before making an http call, similar to the @JsonIgnore annotation in Java?

For instance:

Given this object:

{id: 1, libelle: 'test', mustIgnore: 'test'}

;

I am looking to have it transformed into: {id: 1, libelle: 'test'};

I attempted to use delete object.field, but I'm seeking a more streamlined approach, possibly through annotating my object.

Managing this manually for each object is challenging, especially when dealing with arrays of objects.

UPDATE :

I want to avoid deleting fields individually for every object. Is there a method to specify which fields should be excluded and have it done automatically whenever there's a different object with varying fields to remove?

json angular typescript angular7

Answer 1

Answer №1

Employing ES6 object destructuring allows for obtaining an immutable object excluding a desired property.

const obj = {id: 1, libelle: 'test', mustIgnore: 'test'};

// Destructure to separate mustIgnore and rest properties
const { mustIgnore, ...rest } = obj;

const projectedObject = rest;
 
console.log("old", obj);
console.log("new", projectedObject);

Answer 2

Employing ES6 object destructuring allows for obtaining an immutable object excluding a desired property.

const obj = {id: 1, libelle: 'test', mustIgnore: 'test'};

// Destructure to separate mustIgnore and rest properties
const { mustIgnore, ...rest } = obj;

const projectedObject = rest;
 
console.log("old", obj);
console.log("new", projectedObject);

Answer 3

Answer №2

If you're dealing with an array, a simple way to transform it is by using the map method like this:

const ignoreArray = array
    .map(
        a => ({ id: a.id, label: a.label })
    )

This will give you a new array containing objects with only the id and label properties.

UPDATE: After considering Kunal's suggestion, a more efficient approach would be:

const ignoreArray = array
    .map(
        obj => {
            const { shouldIgnore1, shouldIgnore2, ...remaining } = obj;
            return remaining;
        }
    )

In this scenario, shouldIgnore1, shouldIgnore2, etc. represent the fields that should be omitted from the final result

Answer 4

If you're dealing with an array, a simple way to transform it is by using the map method like this:

const ignoreArray = array
    .map(
        a => ({ id: a.id, label: a.label })
    )

This will give you a new array containing objects with only the id and label properties.

UPDATE: After considering Kunal's suggestion, a more efficient approach would be:

const ignoreArray = array
    .map(
        obj => {
            const { shouldIgnore1, shouldIgnore2, ...remaining } = obj;
            return remaining;
        }
    )

In this scenario, shouldIgnore1, shouldIgnore2, etc. represent the fields that should be omitted from the final result

Answer 5

Answer №3

In situations where your dataset is comprised solely of key+value pairings, a strategy to consider involves initiating an empty object {} and iteratively cycling through all entries within the initial object using a for loop. During this process, you can cross-reference each key against a designated array containing names of keys that should be excluded. If a specific item is found in the exclusion array, it can be bypassed; otherwise, it should be included in the new object being constructed.

Answer 6

In situations where your dataset is comprised solely of key+value pairings, a strategy to consider involves initiating an empty object {} and iteratively cycling through all entries within the initial object using a for loop. During this process, you can cross-reference each key against a designated array containing names of keys that should be excluded. If a specific item is found in the exclusion array, it can be bypassed; otherwise, it should be included in the new object being constructed.

Answer 7

Answer №4

const array = [
{id: 1, libelle: 'test1', mustIgnore: 'ignore1'},
{id: 2, libelle: 'test2', mustIgnore: 'ignore2'},
{id: 3, libelle: 'test3', mustIgnore: 'ignore3'}
];


const freshArray = [];

for (item of array) {
    const { mustIgnore, ...others } = item;
    freshArray.push(others);
}

console.log(freshArray);

Another approach:

const array = [
{id: 1, libelle: 'test1', mustIgnore: 'ignore1'},
{id: 2, libelle: 'test2', mustIgnore: 'ignore2'},
{id: 3, libelle: 'test3', mustIgnore: 'ignore3'}
];

const newArray = array.map(obj => {
                const { mustIgnore, ...rest } = obj;
                return rest;
 });


 console.log(newArray);

For more information, please refer to object destructuring, but this covers the basics!

Answer 8

const array = [
{id: 1, libelle: 'test1', mustIgnore: 'ignore1'},
{id: 2, libelle: 'test2', mustIgnore: 'ignore2'},
{id: 3, libelle: 'test3', mustIgnore: 'ignore3'}
];


const freshArray = [];

for (item of array) {
    const { mustIgnore, ...others } = item;
    freshArray.push(others);
}

console.log(freshArray);

Another approach:

const array = [
{id: 1, libelle: 'test1', mustIgnore: 'ignore1'},
{id: 2, libelle: 'test2', mustIgnore: 'ignore2'},
{id: 3, libelle: 'test3', mustIgnore: 'ignore3'}
];

const newArray = array.map(obj => {
                const { mustIgnore, ...rest } = obj;
                return rest;
 });


 console.log(newArray);

For more information, please refer to object destructuring, but this covers the basics!

Ways to exclude an object field in Angular 7

Answer №1

Answer №2

Answer №3

Answer №4

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