ios - UIPageViewController不同页面的快速加载

Question

我知道如何在 UIPageViewController 中加载与第 1 页不同的页面。问题是我的 UIPageViewController 有 150 页长，当我尝试加载第 100 页时，最多需要 8 秒才能完成该任务。关于如何加快这一过程有什么建议吗？

这是我目前的做法:

let savedPage = self.load(page: bookmarkPageNumber, pageViewController: self)!
self.setViewControllers([savedPage], direction: UIPageViewControllerNavigationDirection.forward, animated: false, completion: nil)

Answer 1

在您的情况下，您应该使用 UIPageViewControllerDataSource ( https://developer.apple.com/reference/uikit/uipageviewcontrollerdatasource ) 而不是 setViewControllers(..)

示例:

yourPageController.dataSource = self

并将此方法添加到包含您的 yourPageController 的类中。不要忘记将此类标记为 UIPageViewControllerDataSource。

    //MARK: - UIPageViewControllerDataSource

    func pageViewController(pageViewController: UIPageViewController,
        viewControllerBeforeViewController viewController: UIViewController) -> UIViewController? {
            guard let viewControllerIndex = orderedViewControllers.indexOf(viewController) else {
                return nil
            }

            let previousIndex = viewControllerIndex - 1

            guard previousIndex >= 0 else {
                return nil
            }

            guard orderedViewControllers.count > previousIndex else {
                return nil
            }

            return orderedViewControllers[previousIndex]
    }

    func pageViewController(pageViewController: UIPageViewController,
        viewControllerAfterViewController viewController: UIViewController) -> UIViewController? {
            guard let viewControllerIndex = orderedViewControllers.indexOf(viewController) else {
                return nil
            }

            let nextIndex = viewControllerIndex + 1
            let orderedViewControllersCount = orderedViewControllers.count

            guard orderedViewControllersCount != nextIndex else {
                return nil
            }

            guard orderedViewControllersCount > nextIndex else {
                return nil
            }

            return orderedViewControllers[nextIndex]
    }