Concatenating Arrays

Two Arrays

 var array1 = [1, 2]; 
 var array2 = [3, 4, 5];

Version ≥ 3

var array3 = array1.concat(array2); // returns a new array

Version ≥ 6

var array3 = [...array1, ...array2]

Results in a new Array:

[1, 2, 3, 4, 5]

Multiple Arrays

var array1 = ["a", "b"], 
    array2 = ["c", "d"], 
    array3 = ["e", "f"], 
    array4 = ["g", "h"];

Version ≥ 3

Provide more Array arguments to array.concat()

var arrConc = array1.concat(array2, array3, array4);

Version ≥ 6 Provide more arguments to []

var arrConc = [...array1, ...array2, ...array3, ...array4]

Results in a new Array:

["a", "b", "c", "d", "e", "f", "g", "h"]

Without Copying the First Array

 var longArray = [1, 2, 3, 4, 5, 6, 7, 8], 
 shortArray = [9, 10];

Version ≥ 3

Provide the elements of shortArray as parameters to push using Function.prototype.apply

longArray.push.apply(longArray, shortArray);

Version ≥ 6 Use the spread operator to pass the elements of shortArray as separate arguments to push

longArray.push(...shortArray)

The value of longArray is now:

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

Note that if the second array is too long (>100,000 entries), you may get a stack overflow error (because of how apply works). To be safe, you can iterate instead:

shortArray.forEach(function (elem) { longArray.push(elem);
});

Array and non-array values

var array = ["a", "b"];

Version ≥ 3

var arrConc = array.concat("c", "d");

Version ≥ 6

var arrConc = [...array, "c", "d"]

Results in a new Array:

["a", "b", "c", "d"]

You can also mix arrays with non-arrays

var arr1 = ["a","b"]; 
var arr2 = ["e", "f"];
var arrConc = arr1.concat("c", "d", arr2);

Results in a new Array:

["a", "b", "c", "d", "e", "f"]