Algorithm for counting the number of pages (can you suggest a better one)

Question

PHP

I dug into one system and was a little crazy about the amount of code that counts the number of pages there (something like pagination). Tons of variables, checks.
I thought about what is so complicated here and wrote my own implementation:

1. <?php
2. /**
3. * Описание алгоритма:
4. * Алгоритм печатает количество страниц, основываясь на трех константах:
5. * 1. Количество элементов(всего)
6. * 2. Желаемое количество элементов на страницу
7. * 3. Максимальное количество дополнений(ситуация, когда на последней странице 1-2 элемента)
8. */
9. define('RESULTS_COUNT', 12349);
10. define('RESULTS_PER_PAGE', 17);
11. define('MAX_ADDITIONS', 7);
12.  
13. $results_per_page = RESULTS_PER_PAGE>0?RESULTS_PER_PAGE:1; // дополнительная проверка на division by zero
14.  
15. if(RESULTS_COUNT<$results_per_page || RESULTS_COUNT==$results_per_page){
16.   exit('1 page for '.RESULTS_COUNT.' results.');
17. }
18.  
19. $exact_pages_count = RESULTS_COUNT / $results_per_page;
20. if (is_int($exact_pages_count)) {
21.   exit($exact_pages_count.' pages for '.RESULTS_COUNT.' results.');
22. }
23. else {
24.   $rough_estimate_pages = intval($exact_pages_count);
25.   $number_of_surplus=RESULTS_COUNT-($results_per_page*$rough_estimate_pages);
26.   $pages_of_surplus=$number_of_surplus<=MAX_ADDITIONS?$rough_estimate_pages:$rough_estimate_pages++;
27.   exit($rough_estimate_pages.' pages for '.RESULTS_COUNT.' results.');
28. }
29. ?>

* This source code was highlighted with Source Code Highlighter.

Perhaps you have a better solution for the same problem? If yes, that would be interesting to see.

Answer 1

% - remainder of division
floor() - round down
$countPages = floor(RESULTS_COUNT/RESULTS_PER_PAGE);
if (RESULTS_COUNT%RESULTS_PER_PAGE >= MAX_ADDITIONS){
$countPages++;
};
return $countPages;

Answer 2

something is not clear, you need to calculate how many pages the received number of results will take with a given number of elements per page?

Answer 3

if(RESULTS_COUNT<$results_per_page || RESULTS_COUNT==$results_per_page)

O_o
There is a comparison operator <=

Answer 4

Oh, just the day before yesterday it was necessary to solve this problem for a Silverlight application (C# language). Here's what happened:
public static IDictionary<int, IEnumerable> DivideItems(IEnumerable sourceItems, int itemsAtPage)
{
var result = new Dictionary<int, IEnumerable>();
var array = sourceItems.ToArray();
var j = 0;
var pageNumber = 1;
while (j < array.Length)
{
var ar = new T[Math.Min(itemsAtPage, array.Length - j)];
for (var i = 0; i < ar.Length; i++)
ar[i] = array[j + i];
j += ar.Length;
result.Add(pageNumber, ar);
pageNumber++;
}
return result;
}
Это Generic-метод, T - любой тип. Функция принимает коллекцию объектов и максимальное число объектов на странице. Возвращает пары <[номер страницы] : [коллекция объектов страницы]>
Похоже, <code/> не совсем точно показывает обобщенные типы, так что вот на всякий случай скриншот:


Answer 5

DeusModus
The Cheese algorithm just solves the problem. Cheese, after all, uses ceil - finding the smallest integer that is greater than the resulting number.

Answer 6

The author, maybe in this system the number of records is not initially considered, for acceleration.
For example, 5 pages are displayed first, you poke on the fifth page, if it is not there, then you are thrown to the very last page that is.