Firebase - How to Get the Key Value in Observeeventtype = Value

Firebase - how to get the key value in observeEventType = Value

In if condition, you need to get allKeys to get "-KKMkpA22PeoHtBAPyKm" ...

    if snapshot.exists() {
        for a in (snapshot.value?.allKeys)!{
            print(a)
        }
    } else {
        print("we don't have that, add it to the DB now")
    }

Getting key value but not the root value of Firebase

To get all room names ...You have to get allKeys like

  snapshot.value.allKeys() 

This will give you ["user1_user2","user1_user3"] i.e your roomnames.

Firebase database query by key value

First go till child "livefeeds" and then "queryOrderedByChild" category and use equal to child like this way

[[[[self.ref child:@"livefeeds"] queryOrderedByChild:@"category"]   
                                   queryEqualToValue:groupId]   
                                    observeEventType:FIRDataEventTypeValue withBlock:^(FIRDataSnapshot * _Nonnull snapshot) {

    if (snapshot.value != [NSNull null]){
        for (NSDictionary *snap in [snapshot.value allValues]) {
             NSLog(@"---> %@",snap);
        }
    }
}];

How to query by key string instead of value in firebase?

This takes a bit of reframing of the problem. Instead of saying "I want the items with this property", think of it as "I want the items with whatever value for this property".

Query firebaseSearchQuery = mUserDatabase.orderByChild("exp/banana").startAt(" ") 

Here the space (" ") is just the first printable character, so we're matching everything.

While this query will work, it requires that you define an index on exp/banana for each user. It's very likely that you'll get better results if you create an inverted index in your JSON to allow this specific lookup. For a longer explanation and an example, see my answer here: Firebase query if child of child contains a value

How get data in Firebase?

Move this to viewDidLoad
FIRDatabaseReference *ref = [[FIRDatabase database] reference];

1. Check whether you configured everything according to the documentation.

2. Check the key names

3. If everything is fine then try changing this
   Try with observeSingleEventOfType instead of observeEventType

   [[[self.ref child:@"buysell"] child:@"users"] observeSingleEventOfType:FIRDataEventTypeValue withBlock:^(FIRDataSnapshot * _Nonnull snapshot) {
        NSDictionary *dict = snapshot.value;
        NSLog(@"%@",dict);
   
   } withCancelBlock:^(NSError * _Nonnull error) {
       NSLog(@"%@", error.localizedDescription);
   }];
   

4. If above 3 not works check your error statement

Firebase - accessing child node when you don't know the value of the parent key - generated by childByAutoId()

I think you're looking for something like:

ref.child("friendRequest")
   .child(FIRAuth.auth()?.currentUser.uid)
   .queryOrdered(byChild: "fromId")
   .queryEqual(toValue: "theUidThatYou'reLookingFor")
   .observeSingleEvent(of: .childAdded, with: { (snapshot) in
        print("\(snapshot.key)")
    })

Firebase with Swift ambiguous use of observeEventType

Just some coding errors

Remove: (snapshot)-> Void

Change: child in snapshot as snapshot is not a sequence, whereas snapshot.children is

I assume you want to store the friends name as a string and name is a key in your structure. So change self.friendsList.append(child.value) to

let name = child.value["name"] as? String
friendsList.append(name!)

Here's the corrected code:

var friendsList = [String]()

ref.observeSingleEventOfType(.Value, withBlock: { snapshot in

    if snapshot.value is NSNull {

    } else {
        for child in snapshot.children {
            let name = child.value["name"] as? String
            friendsList.append(name!)
        }
        print("\(friendsList)")
    }
})

Firebase snapshot.key not returning actual key?

When you run a query at a location, the result will be a list of the matching children. Even if there is only a single matching item, the result will be a list of one child.

You're printing the key of all resulting children. Since there is no single result, the SDK prints the key of the location/collection that you queried: users.

What you're likely looking for is to loop over the matching children and print their keys:

let query = usersRef.queryOrderedByChild("email").queryEqualToValue(email)
query.observeEventType(.Value, withBlock: { snapshot in
    for child in snapshot.children {
        print(child.key)
    }
})

How to read child or filter firebase data by its key value in swift ios

Since you want to execute the same code for each item, you'll want to use .ChildAdded:

ref.queryOrderedByChild("price").observeEventType(.ChildAdded, withBlock: { snapshot in
  if let price = snapshot.value["price"] as? Int {
    println("\(snapshot.key) price at \(price) Dollars ")
    println(snapshot.key)
  }
})

See the page on retrieving data in the Firebase guide for iOS developers for more information and examples.

Update

I ended up using your code in a local xcode and see there are two problems. So all three combined:

1. you are listening for the .Value event, but your block is dealing with a single item at a time. Solution:

   ref.queryOrderedByChild("price")
      .observeEventType(.ChildAdded, withBlock: { snapshot in
   

2. you are listening for the .Value event at the top-level, but you are adding the items under users. Solution:

   ref.childByAppendingPath("users")
      .queryOrderedByChild("price")
      .observeEventType(.ChildAdded, withBlock: { snapshot in
   

3. you are testing whether the price is an Int, but are adding them as strings. Solution:

   var item1     =  ["name": "Alan Turning", "item" : "Red Chair", "price": 100]
   var item2     =  ["name": "Grace Hopper", "item": "Sofa Bed"  , "price": 120]
   var item3     =  ["name": "James Cook"  , "item": "White Desk", "price": 250]
   var item4     =  ["name": "James Cook"  , "item": "Mattress Cal King", "price": 100]
   

With those changes, the code prints out these results for me:

item1 price at 100 Dollars 
item1
item4 price at 100 Dollars 
item4
item2 price at 120 Dollars 
item2
item3 price at 250 Dollars 
item3

---

Related Topics