I'm trying to push multiple elements as one array, but getting an error:
> a = []
[]
> a.push.apply(null, [1,2])
TypeError: Array.prototype.push called on null or undefined
I'm trying to do similar stuff that I'd do in ruby, I was thinking that apply is something like *.
>> a = []
=> []
>> a.push(*[1,2])
=> [1, 2]
You can push multiple elements into an array in the following way
var a = [];
a.push(1, 2, 3);
console.log(a);
array1.push(...array2) that works exactly like array1.push(array2[0], array2[1], array2[2]) except a limitation of number of elements in array2 (about 100,000)Now in ECMAScript2015 (a.k.a. ES6), you can use the spread operator to append multiple items at once:
var arr = [1];
var newItems = [2, 3];
arr.push(...newItems);
console.log(arr);See Kangax's ES6 compatibility table to see what browsers are compatible
a = []; b = []; for (let i = 0; i < 1000000; i++) { b.push(i); } a.push(...(b.slice(0, 130000))); This indeed causes an error for me. "Uncaught RangeError: Maximum call stack size exceeded" The limit may depend on the size of elements in the array, not just the number of elements. That's just a guess. It may depend on the environment. Also just a guess. I haven't tried looking for documentation.
When using most functions of objects with apply or call, the context parameter MUST be the object you are working on.
In this case, you need a.push.apply(a, [1,2]) (or more correctly Array.prototype.push.apply(a, [1,2]))
As one alternative, you can use Array.concat:
var result = a.concat(b);
This would create and return a new array instead of pushing items to the same array. It can be useful if you don't want to modify the source array but rather make a shallow copy of it.
Array.prototype.push.apply(arr1, arr2) is the correct answer, because using arr1.concat(arr2) you are creating a new array.
arr1 = arr1.concat(arr2) is not a big deal and looks much cleaner. Pushing to old array or replacing old array with the new one depends on your needs. If you deal with 10m+ elements pushing to old array will work faster, if you manage small chunks you hardly find any difference in speed. Both options a completely legit.
prototype.push.apply only calls push once. And the distinction above isn't necessary about speed but an in-place operation vs creating a new array. What if I had a method that took an array and was supposed to modify it in-place? The concat method cannot possibly work, even with VisionN's code as it won't modify the variable for the caller of the function.
a = a.concat(b) is still a shorter syntax than Array.prototype.push.apply(arr1, arr2)If you want an alternative to Array.concat in ECMAScript 2015 (a.k.a. ES6, ES2015) that, like it, does not modify the array but returns a new array you can use the spread operator like so:
var arr = [1];
var newItems = [2, 3];
var newerItems = [4, 5];
var newArr = [...arr, ...newItems, ...newerItems];
console.log(newArr);Note this is different than the push method as the push method mutates/modifies the array.
If you want to see if certain ES2015 features work in your browser check Kangax's compatibility table.
You can also use Babel or a similar transpiler if you do not want to wait for browser support and want to use ES2015 in production.
Easier way is
a = []
a.push(1,2,3)
Another way is
a = [...a, 4,5,6]
if you want to create another array
const b = a.concat(7,8,9)
I had the same doubt and in my case, an easier solution worked for me:
let array = []
array.push(1, 2, 4, "string", new Object())
console.log(array)
// logs [ 1, 2, 4, 'string', {} ]
Imagine you have an array of first ten numbers but missing a number, say 6.
You can insert it into the array at the index 5 with the following code
function insert(array, index, obj) {
return [...array.slice(0,index), obj, ...array.slice(index)]
}
let arr = [1,2,3,4,5,7,8,9,0]
arr = insert(arr, 5, 6)
console.log(arr)Pushing multiple objects at once often depends on how are you declaring your array.
This is how I did
//declaration
productList = [] as any;
now push records
this.productList.push(obj.length, obj2.length, items);
any here
ain place ofnull, like this:a.push.apply(a, [1,2])