The lastIndexOf()
method of Array
instances returns the last index at which a given element can be found in the array, or -1 if it is not present. The array is searched backwards, starting at fromIndex
.
The lastIndexOf()
method of Array
instances returns the last index at which a given element can be found in the array, or -1 if it is not present. The array is searched backwards, starting at fromIndex
.
lastIndexOf(searchElement) lastIndexOf(searchElement, fromIndex)
searchElement
Element to locate in the array.
fromIndex
Optional
Zero-based index at which to start searching backwards, converted to an integer.
fromIndex < 0
, fromIndex + array.length
is used.fromIndex < -array.length
, the array is not searched and -1
is returned. You can think of it conceptually as starting at a nonexistent position before the beginning of the array and going backwards from there. There are no array elements on the way, so searchElement
is never found.fromIndex >= array.length
or fromIndex
is omitted, array.length - 1
is used, causing the entire array to be searched. You can think of it conceptually as starting at a nonexistent position beyond the end of the array and going backwards from there. It eventually reaches the real end position of the array, at which point it starts searching backwards through the actual array elements.The last index of searchElement
in the array; -1
if not found.
The lastIndexOf()
method compares searchElement
to elements of the array using strict equality (the same algorithm used by the ===
operator). NaN
values are never compared as equal, so lastIndexOf()
always returns -1
when searchElement
is NaN
.
The lastIndexOf()
method skips empty slots in sparse arrays.
The lastIndexOf()
method is generic. It only expects the this
value to have a length
property and integer-keyed properties.
The following example uses lastIndexOf()
to locate values in an array.
const numbers = [2, 5, 9, 2]; numbers.lastIndexOf(2); // 3 numbers.lastIndexOf(7); // -1 numbers.lastIndexOf(2, 3); // 3 numbers.lastIndexOf(2, 2); // 0 numbers.lastIndexOf(2, -2); // 0 numbers.lastIndexOf(2, -1); // 3
You cannot use lastIndexOf()
to search for NaN
.
const array = [NaN]; array.lastIndexOf(NaN); // -1
The following example uses lastIndexOf
to find all the indices of an element in a given array, using push()
to add them to another array as they are found.
const indices = []; const array = ["a", "b", "a", "c", "a", "d"]; const element = "a"; let idx = array.lastIndexOf(element); while (idx !== -1) { indices.push(idx); idx = idx > 0 ? array.lastIndexOf(element, idx - 1) : -1; } console.log(indices); // [4, 2, 0]
Note that we have to handle the case idx === 0
separately here because the element will always be found regardless of the fromIndex
parameter if it is the first element of the array. This is different from the indexOf()
method.
You cannot use lastIndexOf()
to search for empty slots in sparse arrays.
console.log([1, , 3].lastIndexOf(undefined)); // -1
The lastIndexOf()
method reads the length
property of this
and then accesses each property whose key is a nonnegative integer less than length
.
const arrayLike = { length: 3, 0: 2, 1: 3, 2: 2, 3: 5, // ignored by lastIndexOf() since length is 3 }; console.log(Array.prototype.lastIndexOf.call(arrayLike, 2)); // 2 console.log(Array.prototype.lastIndexOf.call(arrayLike, 5)); // -1
Desktop | Mobile | Server | ||||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
Chrome | Edge | Firefox | Opera | Safari | Chrome Android | Firefox for Android | Opera Android | Safari on IOS | Samsung Internet | WebView Android | Deno | Node.js | ||
lastIndexOf |
1 | 12 | 1.5 | 9.5 | 3 | 18 | 4 | 10.1 | 1 | 1.0 | ≤37 | 1.0 | 0.10.0 |
© 2005–2023 MDN contributors.
Licensed under the Creative Commons Attribution-ShareAlike License v2.5 or later.
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/lastIndexOf