Array value changing

Question

I have 2 arrays - activeDealers and normalDealers.

let normalDealers = [
   { nickname: 'one', inactive: false, dealer: true },
   { nickname: 'two', inactive: false, dealer: false },
   { nickname: 'three', inactive: false, dealer: false }
]

activeDealers = [...normalDealers]

then the normalDealers array should pass through some conditions like this

if (normalDealers.length > 1) {
    const prev = normalDealers.shift();
    normalDealers.push(prev);
}

After this, the normalDealers array will be like this:

[
    { nickname: 'two', inactive: false, dealer: false },
    { nickname: 'three', inactive: false, dealer: false },
    { nickname: 'one', inactive: false, dealer: true }
]

I want to change the normalDealers array's 0-th index obj value to - { nickname: 'two', inactive: false, dealer: true }

[
    { nickname: 'two', inactive: false, dealer: true },
    { nickname: 'three', inactive: false, dealer: false },
    { nickname: 'one', inactive: false, dealer: false }
]

But I want the original order of the array inside the activeDealers with updated dealer status value.

Expecting output:

[
    { nickname: 'one', inactive: false, dealer: false },
    { nickname: 'two', inactive: false, dealer: true },
    { nickname: 'three', inactive: false, dealer: false }
]

Why do you need to change the original order at all? Why are you using `shift`? Why can't you use a `for loop` and mutate the object at a particular index? — Arjis Chakraborty, Jul 07 '23 at 10:32
Does this answer your question? [Is JavaScript a pass-by-reference or pass-by-value language?](https://stackoverflow.com/questions/518000/is-javascript-a-pass-by-reference-or-pass-by-value-language) — Konrad, Jul 07 '23 at 10:57

score 0 · Answer 1 · edited Jul 07 '23 at 11:31

0

You can use the unshift method. The unshift method inserts new elements at the beginning of an array, and the pop method adds new elements to the end of the array, both return the new length of the array.

if (normalDealers.length > 1) {
    const prev = normalDealers.shift();
    normalDealers.unshift(prev);
}

edited Jul 07 '23 at 11:31

Dmitriy Popov

2,150
3
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answered Jul 07 '23 at 10:39

Nuraiym

1

score 0 · Accepted Answer · edited Jul 07 '23 at 11:31

This should work:

const activeDealers = [
    { nickname: 'one', inactive: false, dealer: true },
    { nickname: 'two', inactive: false, dealer: false },
    { nickname: 'three', inactive: false, dealer: false }
];

const normalDealers = [
    { nickname: 'two', inactive: false, dealer: true },
    { nickname: 'three', inactive: false, dealer: false },
    { nickname: 'one', inactive: false, dealer: false }
];

activeDealers.forEach((entryA, indexA) => {
    normalDealers.forEach((entryB, indexB) => {
        if (entryA.nickname === entryB.nickname) {
            activeDealers[indexA] = normalDealers[indexB];
        }
    });
});

console.log(activeDealers);

It just replaces the entry with the corresponding nickname in activeDealers with the entry in normalDealers, assuming that each nickname occurs only once. I'm not sure if that's the best way to achieve what you're aiming for, though. To determine that, you would have to describe what the meaning behind it is. Why are two arrays needed at all?

Array value changing

2 Answers2