UITableView 滾動到底部

Question

我有這行代碼：

tableView.contentOffset = CGPointMake(0.0f, 10000000.0f);

內容大小比10000000.0f小很多，但 UITableView 仍然沒有滾動到底部。 我該怎么做？

Answer 1

滾動到tableViewCell ？

//for instance, you have 15 cells
NSIndexPath *indexPath = [NSIndexPath indexPathForRow:14 inSection:0];
[self.tableView scrollToRowAtIndexPath:indexPath
                      atScrollPosition:UITableViewScrollPositionTop
                              animated:YES];

Answer 2

您可以創建UIScrollView的擴展，並將其用於 UIScrollView、UITableView 或 UICollectionView 的任何實例。

extension UIScrollView {

    func scrollToBottom(animated: Bool) {
        var y: CGFloat = 0.0
        let HEIGHT = self.frame.size.height
        if self.contentSize.height > HEIGHT {
            y = self.contentSize.height - HEIGHT
        }
        self.setContentOffset(CGPointMake(0, y), animated: animated)
    }
}

每當內容大小大於滾動視圖的高度時，它將相應地滾動到正確的位置。 此方法也適用於 AutoLayout。

Answer 3

Swift 5，iOS 12 測試

👇這段代碼非常適合UITableView ，即使有 0 個元素的部分（這會在我在互聯網上看到的所有其他解決方案中崩潰）

extension UITableView {
    func scrollTableViewToBottom(animated: Bool) {
        guard let dataSource = dataSource else { return }

        var lastSectionWithAtLeasOneElements = (dataSource.numberOfSections?(in: self) ?? 1) - 1

        while dataSource.tableView(self, numberOfRowsInSection: lastSectionWithAtLeasOneElements) < 1 {
            lastSectionWithAtLeasOneElements -= 1
        }

        let lastRow = dataSource.tableView(self, numberOfRowsInSection: lastSectionWithAtLeasOneElements) - 1

        guard lastSectionWithAtLeasOneElements > -1 && lastRow > -1 else { return }

        let bottomIndex = IndexPath(item: lastRow, section: lastSectionWithAtLeasOneElements)
        scrollToRow(at: bottomIndex, at: .bottom, animated: animated)
    }
}

本主題中 Kevin 的固定版本，防止無限滾動0項的 tableView：來源： https : //stackoverflow.com/a/59470799/9763761

func scrollToBottom(animated: Bool) {
    guard let dataSource = dataSource else { return }
    var lastSectionWithAtLeastOneElement = (dataSource.numberOfSections?(in: self) ?? 1) - 1
    while dataSource.tableView(self, numberOfRowsInSection: lastSectionWithAtLeastOneElement) < 1 && lastSectionWithAtLeastOneElement > 0 {
        lastSectionWithAtLeastOneElement -= 1
    }
    let lastRow = dataSource.tableView(self, numberOfRowsInSection: lastSectionWithAtLeastOneElement) - 1
    guard lastSectionWithAtLeastOneElement > -1 && lastRow > -1 else { return }
    let bottomIndex = IndexPath(item: lastRow, section: lastSectionWithAtLeastOneElement)
    scrollToRow(at: bottomIndex, at: .bottom, animated: animated)
}

👇此代碼為UIScrollView ，但不工作UITableView有更多的細胞，然后在一個屏幕可以適應，因為蘋果的怪異內部實現的UITableViewController ：

extension UIScrollView {
    func scrollToBottom(animated: Bool) {
        guard contentSize.height > bounds.size.height else { return }

        let bottomOffset = CGPoint(x: 0, y: contentSize.height - bounds.size.height + contentInset.bottom)
        setContentOffset(bottomOffset, animated: true)
    }
}

Answer 4

不幸的是，我還沒有足夠的代表來發表評論，但是 vol 的回答中有 1 個錯誤，當您在具有 0 個項目的 tableview 上調用它時會導致無限的 while 循環，除此之外它工作得很好！ 更正以下代碼：

func scrollToBottom(animated: Bool) {
    guard let dataSource = dataSource else { return }
    var lastSectionWithAtLeastOneElement = (dataSource.numberOfSections?(in: self) ?? 1) - 1
    while dataSource.tableView(self, numberOfRowsInSection: lastSectionWithAtLeastOneElement) < 1 && lastSectionWithAtLeastOneElement > 0 {
        lastSectionWithAtLeastOneElement -= 1
    }
    let lastRow = dataSource.tableView(self, numberOfRowsInSection: lastSectionWithAtLeastOneElement) - 1
    guard lastSectionWithAtLeastOneElement > -1 && lastRow > -1 else { return }
    let bottomIndex = IndexPath(item: lastRow, section: lastSectionWithAtLeastOneElement)
    scrollToRow(at: bottomIndex, at: .bottom, animated: animated)
}

Answer 5

試試這個代碼：

self.tableView.reloadData()
DispatchQueue.main.asyncAfter(deadline: DispatchTime.now()+0.1, execute: {
    let indexPath = IndexPath(row: self.dateSource.count-1, section: 0)
    self.tableView.scrollToRow(at: indexPath, at: UITableViewScrollPosition.bottom, animated: true)
})