paginate method

Iterable<T> paginate({ required int itemsPerPage, required int page, })

Paginates the elements of the list based on the given parameters.

The paginate method allows you to split a list of elements into multiple pages, with each page containing a specified number of items. This can be useful for implementing paginated UIs, such as displaying a limited number of items per page in a list view.

The itemsPerPage parameter specifies the maximum number of items to include on each page. The page parameter specifies the page number (1-based) for which you want to retrieve the items.

The method returns an Iterable<T> that represents the elements on the specified page. Note that the iterable is lazily evaluated, meaning that elements are computed on-demand as you iterate over it.

If the calculated start index for the page exceeds the length of the list, or if the list is empty, the returned iterable will be empty.

Example usage:

final List<int> numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
final int itemsPerPage = 3;

final Iterable<int> firstPage = numbers.paginate(itemsPerPage: itemsPerPage, page: 1);
print(firstPage.toList()); // Output: [1, 2, 3]

final Iterable<int> secondPage = numbers.paginate(itemsPerPage: itemsPerPage, page: 2);
print(secondPage.toList()); // Output: [4, 5, 6]

Implementation

Iterable<T> paginate({required int itemsPerPage, required int page}) sync* {
  final startIndex = (page - 1) * itemsPerPage;
  final endIndex = startIndex + itemsPerPage;

  for (int i = startIndex; i < endIndex && i < length; i++) {
    yield this[i];
  }
}