ios - 如何在 swift 3.0 中从 UITableViewCell 按钮单击添加弹出框转场？

Question

Here i am using popover to a new ViewController from my UITableVIewCell button click but i am getting this error Main.storyboard: Couldn't compile connection: property=anchorView destination=>

func tableView(_ tableView: UITableView, cellForRowAt indexPath: IndexPath) -> UITableViewCell {

    if(tableView == self.table_view_one) {
        var cell  = table_view_one.dequeueReusableCell(withIdentifier: "first_cell", for: indexPath) as! first_cell
       cell.first_view_time_btn.addTarget(self, action: #selector(Iop_gonio_ViewController.someAction), for: .touchUpInside)
       return cell
    }
}

func someAction() {
    self.performSegue(withIdentifier: "first_time_btn", sender: self)
}

Answer 1

尝试

func tableView(_ tableView: UITableView, cellForRowAt indexPath: IndexPath) -> UITableViewCell {
if(tableView == self.table_view_one) {
    var cell  = table_view_one.dequeueReusableCell(withIdentifier: "first_cell", for: indexPath) as! first_cell
   cell.first_view_time_btn.addTarget(self, action: #selector(someAction(sender:)), for: .touchUpInside)
   return cell
}

func someAction(sender:UIButton) {
    self.performSegue(withIdentifier: "first_time_btn", sender: self)
}

但我认为这不是一个好方法，您应该创建一个处理此类交互的协议(protocol)。