Paginate Javascript array
JavascriptArraysPaginationJavascript Problem Overview
I am trying to write a Javascript function that takes an array
, page_size
and page_number
as parameters and returns an array that mimics paginated results:
paginate: function (array, page_size, page_number) {
return result;
}
so for example when:
array = [1, 2, 3, 4, 5],
page size = 2,
page_number = 2,
the function should return: [3, 4]
.
Any ideas would be appreciated.
Javascript Solutions
Solution 1 - Javascript
You can use Array.prototype.slice
and just supply the params for (start, end)
.
function paginate(array, page_size, page_number) {
// human-readable page numbers usually start with 1, so we reduce 1 in the first argument
return array.slice((page_number - 1) * page_size, page_number * page_size);
}
console.log(paginate([1, 2, 3, 4, 5, 6], 2, 2));
console.log(paginate([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], 4, 1));
Solution 2 - Javascript
Here's a solution with reduce()
:
function paginate (arr, size) {
return arr.reduce((acc, val, i) => {
let idx = Math.floor(i / size)
let page = acc[idx] || (acc[idx] = [])
page.push(val)
return acc
}, [])
}
let array = [1, 2, 3, 4, 5]
let page_size = 2
let pages = paginate(array, page_size)
console.log(pages) // all pages
console.log(pages[1]) // second page
It returns an array of pages so you can get a certain page, or loop through all of them.
Solution 3 - Javascript
I saw an example above that did this correctly (kind of) and wanted to expand on it.
This was the example.
function paginate(array, page_size, page_number) {
// human-readable page numbers usually start with 1, so we reduce 1 in the first argument
return array.slice((page_number - 1) * page_size, page_number * page_size);
}
There are a few things wrong with this.
1.) If the page_number
is 0 then it will try and set the starting split at -1 * page_size
which returns an empty array. So the minimum value of the page_number
attr should be 1, never anything less unless you handle that case in the function.
2.) The starting and ending index of the split are the same. Because of this, you get back an empty array. So the split should be:
return array.split(page_number * page_size, page_number * page_size + page_size)
const myArray = [1,2,3,4,5,6,7,8,9,10];
const paginateBad1 = (array, page_size, page_number) => {
return array.slice((page_number - 1) * page_size, page_number * page_size);
};
const paginateBad2 = (array, page_size, page_number) => {
return array.slice(page_number * page_size, page_number * page_size);
};
const paginateGood = (array, page_size, page_number) => {
return array.slice(page_number * page_size, page_number * page_size + page_size);
};
console.log("Bad 1", paginateBad1(myArray, 2, 0));
console.log("Bad 2", paginateBad2(myArray, 2, 1));
console.log("Good", paginateGood(myArray, 2, 1));
Solution 4 - Javascript
Another aproach that you can utilize, is using .filter, look:
const paginate = function (array, index, size) {
// transform values
index = Math.abs(parseInt(index));
index = index > 0 ? index - 1 : index;
size = parseInt(size);
size = size < 1 ? 1 : size;
// filter
return [...(array.filter((value, n) => {
return (n >= (index * size)) && (n < ((index+1) * size))
}))]
}
var array = [
{id: "1"}, {id: "2"}, {id: "3"}, {id: "4"}, {id: "5"}, {id: "6"}, {id: "7"}, {id: "8"}, {id: "9"}, {id: "10"}
]
var transform = paginate(array, 2, 5);
console.log(transform) // [{"id":"6"},{"id":"7"},{"id":"8"},{"id":"9"},{"id":"10"}]
Solution 5 - Javascript
You can use Array.filter() with the help of its second parameter (the index of the current element being processed in the array).
You'll also need the currently selected page and the number of items per page to display, so you can find the minimum and maximum index of the elements needed.
const indexMin = selectedPage * elementsPerPage;
const indexMax = indexMin + elementsPerPage;
const paginatedArray = arrayToPaginate.filter(
(x, index) => index >= indexMin && index < indexMax
);
Updating the selectedPage and/or the elementsPerPage value will allow to return the correct items to display.
Solution 6 - Javascript
The use of Array#slice
is the expected answer.
Here I use Symbol.iterator to create an iterable.
const arr = [1,2,3,4,5,6,7,8,9,10]
function page({arr, pageSize, pageNumber}) {
const start = pageSize*(pageNumber-1)
const end = pageSize*pageNumber
return {
*[Symbol.iterator]() {
for(let i = start; i < arr.length && i < end; i++) {
yield arr[i]
}
}
}
}
console.log([...page({arr, pageSize: 5, pageNumber: 2})])
Solution 7 - Javascript
Here is another variation using Array.from with Array.slice
const paginate = (array, n) => {
const pageSize = Math.ceil(array.length / n);
return Array.from({ length: pageSize }, (_, index) => {
const start = index * n;
return array.slice(start, start + n);
});
};
Solution 8 - Javascript
Hey I'm sorry I'm a bit late but we can use the Array.splice(start, end)
method except this is much simpler
const page = 2
const step = 2;
const start = page * step - step;
const end = start + step;
const array = [1,2,3,4,5,6]
console.log(array.splice(start, end))
Solution 9 - Javascript
function paginate(array, page_size, page_number) {
// human-readable page numbers usually start with 1, so we reduce 1 in the first argument
return array.slice((page_number - 1) * page_size, page_number * page_size);
}
var arr = [1, 2, 3, 4, 5, 6]
const options = {
//page: parseInt(req.query.page) || 1,
page:1,
limit:10
//limit: parseInt(req.query.limit) || 10,
//customLabels: servCustomLabels,
};
let prev_page = 0;
let next_page = 0;
let h_p_p = null;
let h_n_p = null;
let page_count = Math.ceil((arr.length / options.limit));
if (options.page >= page_count ){ // 2 3
next_page = 0;
}
if(options.page >= 1 && options.page < page_count ){
next_page = options.page + 1;
h_n_p = true;
}else{
next_page = 0;
h_n_p = false;
}
if(options.page <= 1 ){
prev_page =0;
h_p_p = false;
}else{
prev_page = options.page -1 ;
h_p_p = true;
}
console.log(paginate(arr, 2, 2));
console.log({paginator: {
totalDocs: arr.length,
perPage: options.limit,
pageCount: page_count,
currentPage: options.page,
//slNo: 2,
hasPrevPage: h_p_p,
hasNextPage: h_n_p,
prev: prev_page,
next: next_page
}})
Solution 10 - Javascript
function paginate(arr, PerPage) {
let map = {};
let startPage = 1;
arr.forEach((current) => {
if (map[startPage] && map[startPage].length < PerPage) {
map[startPage].push(current);
}
if (!map[startPage]) {
map[startPage] = [current];
}
if (map[startPage] && map[startPage].length >= PerPage) {
startPage++;
}
});
return map;
}
Solution 11 - Javascript
The example below is using iter-ops library (I'm the author).
// our inputs...
const array = [1, 2, 3, 4, 5];
const pageSize = 2;
const pageIndex = 1;
The most efficient way is to process an array as an iterable, so you go through it once.
If you never need other pages, then the fastest way is like this:
import {pipe, skip, page} from 'iter-ops';
const p = pipe(
array,
skip(pageSize * pageIndex), // skip pages we don't want
page(pageSize) // create the next page
).first;
console.log(p); //=> [3, 4]
And if you do need other pages, then you can do:
const p = pipe(
array,
page(pageSize), // get all pages
skip(pageIndex) // skip pages we don't want
).first;
console.log(p); //=> [3, 4]
And in case you need to do further processing:
const i = pipe(
array,
page(pageSize), // get all pages
skip(pageIndex), // skip pages we don't want
take(1), // take just one page
// and so on, you can process it further
);
console.log([...i]); //=> [[3, 4]]
Solution 12 - Javascript
A simple solution using filter
:
function paginate(array, pageIndex, pageSize) {
const first = pageIndex * pageSize
const last = (pageIndex * pageSize) + pageSize
return array.filter((_, index) => {
return first <= index && index < last
})
}