IOS Firebase查询按值访问

Question

如何使用查询按值获取json对象。在详细信息中，想要获取id {4的1而不循环json表中的所有事件

Answer 1

Swift 2

FIRDatabase.database().reference().child("events").queryOrderedByChild("id").queryEqualToValue(4).observeSingleEventOfType(.Value, withBlock: {(Snap) in 
  if let snapDict = Snap.value as? [String:AnyObject]{
      for each in snapDict{
          print(each.0) // Will print out your node ID = 1
          print(each.1) //Will print out your Dictionary inside that node.
          }

      }

  })

Swift 3

 FIRDatabase.database().reference().child("events").queryOrdered(byChild: "id").queryEqual(toValue: 4).observeSingleEvent(of: .value, with: {(Snap) in
        if let snapDict = Snap.value as? [String:AnyObject]{
            for each in snapDict{
                print(each.0) // Will print out your node ID = 1
                print(each.1) //Will print out your Dictionary inside that node.
            }

        }

    })

Answer 2

可能会解决您的问题

[self.dbRef observeEventType:FIRDataEventTypeValue withBlock:^(FIRDataSnapshot * _Nonnull snapshot) {
    if (![snapshot.value isKindOfClass:[NSNull class]]) {
        NSArray *arrResponse = (NSArray *)snapshot.value;

        NSPredicate *pred = [NSPredicate predicateWithFormat:@"id = %@", serchID];
        NSArray *arrSearchedVal = [arrResponse filteredArrayUsingPredicate:pred];
    }
}];

NB： dbRef是您的FIRDatabaseReference，它返回整个数据JSON。然后将NSPredicate与您的搜索ID一起使用，您将获得结果。

快乐编码..