通过观察Firebase中的数据库引用无法获取顶级节点的值

Question

我试图从Swift语言的Firebase数据库中获取分支列表（在本示例中为[branch1，branch2]），当我观察到顶部节点（“ branch”）时，它什么也不返回，甚至第一个print函数都没有被调用，但是如果我观察到像branch1这样的分支之一，则可以正常工作。 现在如何检索分支列表（[branch1，branch2]）

    let ref = FIRDatabase.database().reference()

    ref.child("branches").observe(.value, with: { snapshot in

        print("+++++++++++++++++ ")

        print("++++++++++++\(snapshot.hasChildren())")
        for child in snapshot.children{

            let branch = child as! FIRDataSnapshot
            let branchName = branch.key as! String
            print("\(branchName)") 
       }

   }

在此处输入图片说明

{  "Users" : {
"6dICrkbVVJWzVZ00Vq4y9Fm2Qcg2" : {
  "branches" : {
    "branch1" : true
  },
  "isAdmin" : true
},
"M0C1XUyboPg5zbgQi24nh3SlJri1" : {
  "branches" : {
    "branch2" : true
  }
}  },"branches" : {
"branch1" : {
  "address" : "NDG",
  "campaigns" : {
    "valentine's day" : true
  },
  "manager" : "y",
  "rate" : {
    "average" : 0,
    "bad" : 0,
    "good" : 0
  },
  "users" : {
    "6dICrkbVVJWzVZ00Vq4y9Fm2Qcg2" : true,
    "M0C1XUyboPg5zbgQi24nh3SlJri1" : {
      "isAdmin" : true
    }
  }
},
"branch2" : {
  "address" : "downtown montreal",
  "manager" : "x",
  "rate" : {
    "average" : 0,
    "bad" : 0,
    "good" : 0
  },
  "users" : {
    "M0C1XUyboPg5zbgQi24nh3SlJri1" : true,
    "Z6b5WalkNWZLMVdwGPHi3vSvSRU2" : {
      "isAdmin" : true
    }
  }
}  },  "campaigns" : {
"valentine's day" : {
  "branches" : {
    "branch1" : true
  }
}  }}

和获得branch1的工作代码是：

   ref.child("branches").child("branch1").observe(.value, with: { snapshot in

      let address = snapshot.childSnapshot(forPath: "address").value 
      print("\(address)")
   }

它将打印正确的结果（NDG）

安全规则：

{  "rules": {
"Users":{
  "$uid":{
    ".read":"auth.uid == $uid || root.child('Users').child(auth.uid).child('isAdmin').val()==true",
    ".write":"auth.uid == $uid || root.child('Users').child(auth.uid).child('isAdmin').val()==true"


  }

},  "branches":{
"$branchId":{
".read" : "auth.uid == true || root.child('Users').child(auth.uid).child('isAdmin').val()==true || root.child('branches').child($branchId).child('users').child(auth.uid).exists()",
        ".write" :  "auth.uid == true || root.child('Users').child(auth.uid).child('isAdmin').val()==true || root.child('branches').child($branchId).child('users').child(auth.uid).child('isAdmin').val()==true"

}}}}

Answer 1

感谢@vzsg，我发现问题出在权限和规则上。 所以我改变了规则，很容易得到结果

{  "rules": {
      "Users":{
        "$uid":{
    ".read":"auth.uid == $uid ||       root.child('Users').child(auth.uid).child('isAdmin').val()==true",
    ".write":"auth.uid == $uid || root.child('Users').child(auth.uid).child('isAdmin').val()==true"


  }

},  "branches":{
        ".read": " root.child('Users').child(auth.uid).child('isAdmin').val()==true" ,
        ".write" :  "root.child('Users').child(auth.uid).child('isAdmin').val()==true ",

"$branchId":{
   ".read": " root.child('branches').child($branchId).child('users').child(auth.uid).exists()",
   ".write" :  "root.child('branches').child($branchId).child('users').child(auth.uid).child('isAdmin').val()==true"

}}  }}