How to add a value in an Array?

Question

Let's imagine I have this :

var fruits = [{ "Banana" : "yellow", "lemon" : "yellow"}, {"Orange" : "orange", "Mango" : "orange"}, {"Apple" : "red", "strawberry" : "red"}];

fruits.push({"Kiwi" : "green", "Green Apple" : "green", "watermelon" : "green" });

console.log(fruits);

Now I have an Array "fruits", with objects inside, and I'm using "push()" method to add new objects.

If I would like to add a new yellow fruit, like "Jamaican Banana : yellow" inside the yellow things object, how could I do that ?

Your example isn’t syntactically valid. If you meant {"Banana": "yellow", "lemon": "yellow"}… why store them like that? It’s the least useful structure. {"yellow": ["Banana", "lemon"], "orange": …}" would make a lot more sense. — Ry-
– Ry- ♦, Commented Mar 17, 2017 at 7:45
Actually, it's just an random exemple. In my project I'm using a firebase ref to fill up the array using a foreach. Then I have my Array, fill up with objects inside that I transfer to the scope. And I would like to know if it exist a simple way to add something in these objects. Because the problem with the "push" method is, each push create a new object. Wich is an action I want, but I also would like to add something more in each steps of my loop, am I enough clear ? — Memphis
– Memphis, Commented Mar 17, 2017 at 7:50

mehulmpt · Accepted Answer · 2017-03-17 07:47:52Z

3

What you're trying to do won't work very well. Try something like this:

var fruits = {
  yellow: ["fruit1", "2", "3"],
  orange: ["4","5","6"]
};

// later
console.log(fruits);
fruits.yellow.push("another fruit");
console.log(fruits);

answered Mar 17, 2017 at 7:47

mehulmpt

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Comments

Nina Scholz · Accepted Answer · 2017-03-17 07:46:56Z

2

You need to quote the keys, if necessary and the string values, but not the wole part with colon inside: Object initializer.

var fruits = [{ Banana: "yellow", lemon: "yellow" }, { Orange: "orange", Mango: "orange" }, { Apple: "red", strawberry: "red" }];

fruits.push({ Kiwi: "green", "Green Apple": "green", watermelon: "green"});

console.log(fruits);

answered Mar 17, 2017 at 7:46

Nina Scholz

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Comments

Community · Accepted Answer · 2017-05-23 11:47:06Z

2

Use dynamic property accesses (bracket notation), and don't forget the quotes that go between property names and : in your other fruit objects:

var fruits = [{
  "Banana": "yellow",
  "Lemon": "yellow"
}, {
  "Orange": "orange",
  "Mango": "orange"
}, {
  "Apple": "red",
  "Strawberry": "red"
}]

fruits.push({
  "Kiwi": "green",
  "Green Apple": "green",
  "Watermelon": "green"
})

fruits[0]["Jamaican Banana"] = "yellow"

console.log(fruits)

However, I would suggest (like Ryan and Mehul Mohan have already) using arrays within a top-level fruits object instead (in other words, sorting your data by color):

var fruits = {
  "yellow": ["Banana", "Lemon"],
  "orange": ["Orange", "Mango"],
  "red": ["Apple", "Strawberry"]
}

fruits["green"] = ["Kiwi", "Green Apple", "Watermelon"]

fruits["yellow"].push("Jamaican Banana")

console.log(fruits)

edited May 23, 2017 at 11:47

CommunityBot

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answered Mar 17, 2017 at 7:48

gyre

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3 Comments

Memphis Over a year ago

Oh that's seems to fit to my needs ! How would you do the same in a same Array size loop ? Let's imagine we have "fruits[iteration][fruit] = color", "iteration", "fruit" and "color" as variables.

gyre Over a year ago

Just like that: for (var i = 0; i < fruits.length; i++) { fruits[i]["Example Fruit"] = "example-color" }. However, this will mess up the sorting of your data which is why the alternative data structure I suggested is more desirable.

Memphis Over a year ago

Actually, it's just an random exemple. In my project I'm using a firebase ref to fill up the array using a foreach. Then I have my Array, fill up with objects inside that I transfer to the scope. And I would like to know if it exist a simple way to add something in these objects. Because the problem with the "push" method is, each push create a new object. Wich is an action I want, but I also would like to add something more in each steps of my loop, am I enough clear ?

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