ios - 使用Prepareforsegue将数据传递给发现nil的segued Controller

Question

我要继续上以下类(class)。

class FriendsProfileViewController: UIViewController {
    var user : User!

    override func viewDidLoad() {
        super.viewDidLoad()
        print(user.firstName)
    }
}

在 segueing 类上，我执行了以下操作来传递数据

var friends: [User]?

override func tableView(tableView: UITableView, didSelectRowAtIndexPath indexPath: NSIndexPath) {
    performSegueWithIdentifier("viewFriend", sender: indexPath)
}

override func prepareForSegue(segue: UIStoryboardSegue, sender: AnyObject?) {
    let destinationVC = segue.destinationViewController
    if let friendsProfileVC = destinationVC as? FriendsProfileViewController{
        if let indexPath = sender as? NSIndexPath{
            print(indexPath.row)
            friendsProfileVC.user = friends?[indexPath.row]
        }
    }
}

为什么我会收到错误

fatal error: unexpectedly found nil while unwrapping an Optional value

即使我的friend?[indexPath.row]在我segue?时被验证包含一个值。

Answer 1

尝试在执行 segue 之前获取用户并将其作为 segue 发送者传递。

var friends: [User]?

private func friend(at index: Int) -> User? {

    return friends.flatMap { $0.indicies.contains(index) ? $0[index] : nil }
}

override func tableView(tableView: UITableView, didSelectRowAtIndexPath indexPath: NSIndexPath) {

    if let friend = friend(at: indexPath.row) {
        performSegueWithIdentifier("viewFriend", sender: friend)
    } else {
        print("Invalid index: \(indexPath.row)")
    }
}

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

    switch (segue.destinationViewController, sender) {
    case (let controller as FriendsProfileViewController, let user as User):
        controller.user = user
    default:
        break
    }
}