ios - 如何从 UITableView 滑动触发 segue 以快速编辑

Question

我正在尝试在我的 UITableView 中包含滑动功能，我想从那里执行一个 segue。例如，当单元格向左滑动时，它会显示一个 Edit 按钮，当按下这个编辑按钮时，它会打开一个新的 ViewController.UI

Answer 1

试试下面的代码:

let cSelector : Selector = “swipeAction”

let rightSwipe = UISwipeGestureRecognizer(target: self, action: cSelector)
rightSwipe.direction = UISwipeGestureRecognizerDirection.Right
mainTableView.addGestureRecognizer(rightSwipe) // Or you can add it to cell

并创建滑动 Action :

func swipeAction()
{
    // TODO 
    self.performSegueWithIdentifier("segueId", sender: nil)
}

或者您可以在滑动单元格时直接调用 segue(如果您实现了像 canEditCellForRow 这样的内置滑动)

并实现prepareForSegue方法

// MARK: - Navigation

// In a storyboard-based application, you will often want to do a little preparation before navigation

override func prepareForSegue(segue: UIStoryboardSegue, sender: AnyObject?) {

    // Get the new view controller using segue.destinationViewController.
    // Pass the selected object to the new view controller.

    let controller = segue.destinationViewController as YourDestViewController
    controller.someData = // Pass some object if you need
}