JS: merging 2 arrays within an object

Question

Given a JSON object, how would you merge the batter array and topping array within the object. I know I can do let x = foo.batters.batter.concat(foo.topping), which will give me an array of the merged array but what if I wanted it within object?

foo = {
    "id": "0001",
    "type": "donut",
    "name": "Cake",
    "ppu": 0.55,
    "batters":
        {
            "batter":
                [
                    { "id": "1001", "type": "Regular" },
                    { "id": "1002", "type": "Chocolate" },
                    { "id": "1003", "type": "Blueberry" },
                    { "id": "1004", "type": "Devil's Food" }
                ]
        },
    "topping":
        [
            { "id": "5001", "type": "None" },
            { "id": "5002", "type": "Glazed" },
            { "id": "5005", "type": "Sugar" },
            { "id": "5007", "type": "Powdered Sugar" },
            { "id": "5006", "type": "Chocolate with Sprinkles" },
            { "id": "5003", "type": "Chocolate" },
            { "id": "5004", "type": "Maple" }
        ]
}

either concat it to batter array or topping array -- either or, just curious as to how to do it — DaBosco
– DaBosco, Commented Dec 7, 2020 at 16:37
where ? in foo.batters.batter or in foo.topping or in foo.newOne, ? , and what will happent to previous lists ? — Mister Jojo
– Mister Jojo, Commented Dec 7, 2020 at 16:38
foo.batters.batter = foo.batters.batter.concat(foo.topping) — Wais Kamal
– Wais Kamal, Commented Dec 7, 2020 at 16:38

Tarık Çalışkan · Accepted Answer · 2020-12-07 19:26:56Z

2

Your answer is partially true. But you didn't add the value to the foo object.

You can do it like this:

foo.mergedArray = foo.batters.batter.concat(foo.topping);

After that, when you call foo.mergedArray you can access merged array of batter and topping.

If one of the batter and topping has changed foo.mergedArray will not be updated. Because foo.mergedArray has its own referance.

edited Dec 7, 2020 at 19:26

answered Dec 7, 2020 at 19:21

Tarık Çalışkan

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Or Assayag · Accepted Answer · 2020-12-07 18:01:05Z

1

Try this:

foo.batters.batter = [...foo.batters.batter, ...foo.topping];

answered Dec 7, 2020 at 18:01

Or Assayag

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