I have var ar = [1, 2, 3, 4, 5] and want some function getSubarray(array, fromIndex, toIndex), that result of call getSubarray(ar, 1, 3) is new array [2, 3, 4].
5 Answers
Take a look at Array.slice(begin, end)
const ar = [1, 2, 3, 4, 5];
// slice from 1..3 - add 1 as the end index is not included
const ar2 = ar.slice(1, 3 + 1);
console.log(ar2);For a simple use of slice, use my extension to Array Class:
Array.prototype.subarray = function(start, end) {
if (!end) { end = -1; }
return this.slice(start, this.length + 1 - (end * -1));
};
Then:
var bigArr = ["a", "b", "c", "fd", "ze"];
Test1:
bigArr.subarray(1, -1);
< ["b", "c", "fd", "ze"]
Test2:
bigArr.subarray(2, -2);
< ["c", "fd"]
Test3:
bigArr.subarray(2);
< ["c", "fd","ze"]
Might be easier for developers coming from another language (i.e. Groovy).
const array_one = [11, 22, 33, 44, 55];
const start = 1;
const end = array_one.length - 1;
const array_2 = array_one.slice(start, end);
console.log(array_2);I have var ar = [1, 2, 3, 4, 5] and want some function getSubarray(array, fromIndex, toIndex), that result of call getSubarray(ar, 1, 3) is new array [2, 3, 4].
Exact Solution
function getSubarray(array, fromIndex, toIndex) {
return array.slice(fromIndex, toIndex+1);
}
Let's Test the Solution
let ar = [1, 2, 3, 4, 5]
getSubarray(ar, 1, 3)
// result: [2,3,4]
