ios - Swift 4：从表视图传递到视图控制器

问题描述

我花了几个小时查看相同的问题，但我找到的答案都没有解决这个问题。需要来自 Firebase 数据库的数据从 tableview 传递到另一个视图控制器。需要显示的 PD 数据不在单元格中，我需要从单击表单视图行的 ID 从 firebase 获取。我已经和手写的照片来解释我需要什么。

这是我来自 VC 1 的代码：

func tableView(_ tableView: UITableView, numberOfRowsInSection section: Int) -> Int {
    return guestsList.count
}

func tableView(_ tableView: UITableView, cellForRowAt indexPath: IndexPath) -> UITableViewCell {
    let cell = tableView.dequeueReusableCell(withIdentifier: "guestCell", for: indexPath) as! GuestTableViewCell
    let guest: GuestModel
    guest = guestsList[indexPath.row]
    
    cell.guestNameLabel.text = guest.guestName
    cell.attendingLabel.text = guest.attendingStatus
    cell.menuLabel.text = guest.menuStatus
    cell.ageLabel.text = guest.ageStatus
    cell.tableLabel.text = guest.tableNo
    
    return cell
}

func tableView(_ tableView: UITableView, didSelectRowAt indexPath: IndexPath) {
    let selectedGuest = guestsList[indexPath.row]
    
    let controller = self.storyboard?.instantiateViewController(identifier: "GuestDetail") as! GuestDetailsViewController
    
    controller.guestUser = selectedGuest
   self.present(controller, animated: true, completion: nil)

}

func tableView(_ tableView: UITableView, commit editingStyle: UITableViewCell.EditingStyle, forRowAt indexPath: IndexPath) {
    
    if editingStyle == .delete {
        let guest = guestsList[indexPath.row]
        self.guestsList.remove(at: indexPath.row)
        self.guestsTableView.reloadData()
        self.deleteGuest(id: guest.id!)
    }
}

和 VC 2

var ref: DatabaseReference!
let uid = Auth.auth().currentUser?.uid
var guestDetail = [GuestModel]()
var guestUser: GuestModel?
var documenID: String?



override func viewDidLoad() {
    super.viewDidLoad()
    
  showGuestDetails()
  
}

func showGuestDetails(){
    ref = Database.database().reference().child("userInfo").child(uid!).child("guests")
    ref.queryOrderedByKey().observeSingleEvent(of: .value) { (snapshot) in
        
        if snapshot.childrenCount>0{
            self.guestDetail.removeAll()

            for guests in snapshot.children.allObjects as![DataSnapshot]{
                let guestObject = guests.value as? [String: AnyObject]
                let name = guestObject?["guestName"]
                let familyName = guestObject?["guestFamilyName"]
                let phone = guestObject?["guestPhoneNumber"]
                let email = guestObject?["guestEmail"]

                let guest = GuestModel(guestName: name as? String, guestFamilyName: familyName as! String, guestPhoneNumber: phone as? String, guestEmail: email as? String)


                self.phoneNoLabel.text = guest.guestPhoneNumber
                self.emailLabel.text = guest.guestEmail
            }
        }
    }
    self.nameLabel.text = guestUser!.guestName
}'''

标签： iosswiftuitableviewfirebase-realtime-database

您可以通过 2 种方式达到预期的效果：

将 autoId 从 1st VC 传递到 2nd Detail VC，并showGuestDetails()作为参考

ref = Database.database().reference().child("userInfo").child(uid!).child("guests").child(passedAutoID)

保持相同的引用，但使用 if 条件检查用户是否与要求的相同

for guests in snapshot.children.allObjects as![DataSnapshot]{
     let guestObject = guests.value as? [String: AnyObject]
     guard let email = guestObject?["guestEmail"] as? String,
           email != guestUser?.email else {
           continue
     }

     let name = guestObject?["guestName"]
     let familyName = guestObject?["guestFamilyName"]
     let phone = guestObject?["guestPhoneNumber"

     let guest = GuestModel(guestName: name as? String, guestFamilyName: familyName as! String, guestPhoneNumber: phone as? String, guestEmail: email as? String)
     self.nameLabel.text = guest.guestName
     self.phoneNoLabel.text = guest.guestPhoneNumber
     self.emailLabel.text = guest.guestEmail
 }

ios - Swift 4：从表视图传递到视图控制器 - Firebase？

问题描述

解决方案

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