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0

I created an iOS app that stores the data to Firebase realtime database like this:

users
- ID
-- Email: "test@domain.com"
-- UsersName: "test"
-- Level: 1
-- XP: 0

Now I started with Android development using Kotlin and the data is stored with the first letter of the values name lowercased like this:

users
- ID
-- email: "test@domain.com"
-- usersName: "test" (No mistake, the N is uppercased)
-- level: 1
-- xp: 0 (No mistake, the P is lowercase)

This is my code:

data class User(var UsersName: String = "",
 var XP: Int = 0,
 var Level: Int = 1,
 var Email: String = "")

private fun saveUserToFirebaseDatabase()
 val uid = FirebaseAuth.getInstance().uid ?: ""
 val ref = FirebaseDatabase.getInstance().getReference("users/$uid")
 val mDatabase = FirebaseDatabase.getInstance().getReference();
 val username = usernameSignUp.text.toString()
 val user = User(username, 0, 1, SignUpEmailtext.text.toString())
 print(user)
 mDatabase.child("users").child(uid).setValue(user)
 .addOnSuccessListener 

 
 .addOnFailureListener

 

 

I expect the names of the values to be exactly how they are named in the class (like the iOS app saves them).

android firebase firebase-realtime-database kotlin

share|improve this question

edited Mar 29 at 11:03

Alex Mamo

57.8k1111 gold badges4242 silver badges7878 bronze badges

asked Mar 28 at 17:38

Alex KarapanosAlex Karapanos

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add a comment
 | 

0

I created an iOS app that stores the data to Firebase realtime database like this:

users
- ID
-- Email: "test@domain.com"
-- UsersName: "test"
-- Level: 1
-- XP: 0

Now I started with Android development using Kotlin and the data is stored with the first letter of the values name lowercased like this:

users
- ID
-- email: "test@domain.com"
-- usersName: "test" (No mistake, the N is uppercased)
-- level: 1
-- xp: 0 (No mistake, the P is lowercase)

This is my code:

data class User(var UsersName: String = "",
 var XP: Int = 0,
 var Level: Int = 1,
 var Email: String = "")

private fun saveUserToFirebaseDatabase()
 val uid = FirebaseAuth.getInstance().uid ?: ""
 val ref = FirebaseDatabase.getInstance().getReference("users/$uid")
 val mDatabase = FirebaseDatabase.getInstance().getReference();
 val username = usernameSignUp.text.toString()
 val user = User(username, 0, 1, SignUpEmailtext.text.toString())
 print(user)
 mDatabase.child("users").child(uid).setValue(user)
 .addOnSuccessListener 

 
 .addOnFailureListener

 

 

I expect the names of the values to be exactly how they are named in the class (like the iOS app saves them).

android firebase firebase-realtime-database kotlin

share|improve this question

edited Mar 29 at 11:03

Alex Mamo

57.8k1111 gold badges4242 silver badges7878 bronze badges

asked Mar 28 at 17:38

Alex KarapanosAlex Karapanos

12111 gold badge22 silver badges99 bronze badges

add a comment
 | 

I created an iOS app that stores the data to Firebase realtime database like this:

users
- ID
-- Email: "test@domain.com"
-- UsersName: "test"
-- Level: 1
-- XP: 0

Now I started with Android development using Kotlin and the data is stored with the first letter of the values name lowercased like this:

users
- ID
-- email: "test@domain.com"
-- usersName: "test" (No mistake, the N is uppercased)
-- level: 1
-- xp: 0 (No mistake, the P is lowercase)

This is my code:

data class User(var UsersName: String = "",
 var XP: Int = 0,
 var Level: Int = 1,
 var Email: String = "")

private fun saveUserToFirebaseDatabase()
 val uid = FirebaseAuth.getInstance().uid ?: ""
 val ref = FirebaseDatabase.getInstance().getReference("users/$uid")
 val mDatabase = FirebaseDatabase.getInstance().getReference();
 val username = usernameSignUp.text.toString()
 val user = User(username, 0, 1, SignUpEmailtext.text.toString())
 print(user)
 mDatabase.child("users").child(uid).setValue(user)
 .addOnSuccessListener 

 
 .addOnFailureListener

 

 

I expect the names of the values to be exactly how they are named in the class (like the iOS app saves them).

android firebase firebase-realtime-database kotlin

share|improve this question

edited Mar 29 at 11:03

Alex Mamo

57.8k1111 gold badges4242 silver badges7878 bronze badges

asked Mar 28 at 17:38

Alex KarapanosAlex Karapanos

12111 gold badge22 silver badges99 bronze badges

users
- ID
-- Email: "test@domain.com"
-- UsersName: "test"
-- Level: 1
-- XP: 0

users
- ID
-- email: "test@domain.com"
-- usersName: "test" (No mistake, the N is uppercased)
-- level: 1
-- xp: 0 (No mistake, the P is lowercase)

data class User(var UsersName: String = "",
 var XP: Int = 0,
 var Level: Int = 1,
 var Email: String = "")

private fun saveUserToFirebaseDatabase()
 val uid = FirebaseAuth.getInstance().uid ?: ""
 val ref = FirebaseDatabase.getInstance().getReference("users/$uid")
 val mDatabase = FirebaseDatabase.getInstance().getReference();
 val username = usernameSignUp.text.toString()
 val user = User(username, 0, 1, SignUpEmailtext.text.toString())
 print(user)
 mDatabase.child("users").child(uid).setValue(user)
 .addOnSuccessListener 

 
 .addOnFailureListener

 

 

android firebase firebase-realtime-database kotlin

android firebase firebase-realtime-database kotlin

share|improve this question

edited Mar 29 at 11:03

Alex Mamo

57.8k1111 gold badges4242 silver badges7878 bronze badges

asked Mar 28 at 17:38

Alex KarapanosAlex Karapanos

12111 gold badge22 silver badges99 bronze badges

share|improve this question

edited Mar 29 at 11:03

Alex Mamo

57.8k1111 gold badges4242 silver badges7878 bronze badges

asked Mar 28 at 17:38

Alex KarapanosAlex Karapanos

12111 gold badge22 silver badges99 bronze badges

edited Mar 29 at 11:03

Alex Mamo

57.8k1111 gold badges4242 silver badges7878 bronze badges

edited Mar 29 at 11:03

Alex Mamo

57.8k1111 gold badges4242 silver badges7878 bronze badges

asked Mar 28 at 17:38

Alex KarapanosAlex Karapanos

12111 gold badge22 silver badges99 bronze badges

asked Mar 28 at 17:38

Alex KarapanosAlex Karapanos

12111 gold badge22 silver badges99 bronze badges

1 Answer
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0

Kotlin works on JVM (Java Virtual Machine), so in order to be able to run a .kt file, first of all it should be compiled into a Java file. So when using a class that looks like:

data class User(var UsersName: String = "",
 var XP: Int = 0,
 var Level: Int = 1,
 var Email: String = "")

The corresponding Java file is created and according to the Java Naming Conventions regading variables:

It should be a lowercase letter such as java, lang.

This means that doesn't matter if you choose the name of your fields to start either with an uppercase or a lowercase, all fiels are converted to start with a lowercase. That's why the properties in your database are present like that. Not only in Java/Kotlin is this convention available but also in other programming languages, so I recommend you to use this recommendation in your projects.

Regarding Kotlin, remember that at the end, everything is converted to byte code so it can work on JVM.

share|improve this answer

answered Mar 29 at 11:01

Alex MamoAlex Mamo

57.8k1111 gold badges4242 silver badges7878 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Okay, this means that I have to change the entire iOS app. That's nearly impossible.. My database has a lot of users already.. I wish I new that before I started with iOS version.. Is there any other solution?
  
  – Alex Karapanos
  Mar 29 at 11:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  The rules of naming the variables are also specified in the official documentation. I don't know if you can force Kotlin not to do that. I think you should ask a separate question for that. Regarding reading those kind of fields, you can take a look at my answer from this post.
  
  – Alex Mamo
  Mar 29 at 11:27
  
  

  
  
  
  

add a comment
 | 

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0

Kotlin works on JVM (Java Virtual Machine), so in order to be able to run a .kt file, first of all it should be compiled into a Java file. So when using a class that looks like:

data class User(var UsersName: String = "",
 var XP: Int = 0,
 var Level: Int = 1,
 var Email: String = "")

The corresponding Java file is created and according to the Java Naming Conventions regading variables:

It should be a lowercase letter such as java, lang.

This means that doesn't matter if you choose the name of your fields to start either with an uppercase or a lowercase, all fiels are converted to start with a lowercase. That's why the properties in your database are present like that. Not only in Java/Kotlin is this convention available but also in other programming languages, so I recommend you to use this recommendation in your projects.

Regarding Kotlin, remember that at the end, everything is converted to byte code so it can work on JVM.

share|improve this answer

answered Mar 29 at 11:01

Alex MamoAlex Mamo

57.8k1111 gold badges4242 silver badges7878 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Okay, this means that I have to change the entire iOS app. That's nearly impossible.. My database has a lot of users already.. I wish I new that before I started with iOS version.. Is there any other solution?
  
  – Alex Karapanos
  Mar 29 at 11:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  The rules of naming the variables are also specified in the official documentation. I don't know if you can force Kotlin not to do that. I think you should ask a separate question for that. Regarding reading those kind of fields, you can take a look at my answer from this post.
  
  – Alex Mamo
  Mar 29 at 11:27
  
  

  
  
  
  

add a comment
 | 

0

Kotlin works on JVM (Java Virtual Machine), so in order to be able to run a .kt file, first of all it should be compiled into a Java file. So when using a class that looks like:

data class User(var UsersName: String = "",
 var XP: Int = 0,
 var Level: Int = 1,
 var Email: String = "")

The corresponding Java file is created and according to the Java Naming Conventions regading variables:

It should be a lowercase letter such as java, lang.

This means that doesn't matter if you choose the name of your fields to start either with an uppercase or a lowercase, all fiels are converted to start with a lowercase. That's why the properties in your database are present like that. Not only in Java/Kotlin is this convention available but also in other programming languages, so I recommend you to use this recommendation in your projects.

Regarding Kotlin, remember that at the end, everything is converted to byte code so it can work on JVM.

share|improve this answer

answered Mar 29 at 11:01

Alex MamoAlex Mamo

57.8k1111 gold badges4242 silver badges7878 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Okay, this means that I have to change the entire iOS app. That's nearly impossible.. My database has a lot of users already.. I wish I new that before I started with iOS version.. Is there any other solution?
  
  – Alex Karapanos
  Mar 29 at 11:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  The rules of naming the variables are also specified in the official documentation. I don't know if you can force Kotlin not to do that. I think you should ask a separate question for that. Regarding reading those kind of fields, you can take a look at my answer from this post.
  
  – Alex Mamo
  Mar 29 at 11:27
  
  

  
  
  
  

add a comment
 | 

Kotlin works on JVM (Java Virtual Machine), so in order to be able to run a .kt file, first of all it should be compiled into a Java file. So when using a class that looks like:

data class User(var UsersName: String = "",
 var XP: Int = 0,
 var Level: Int = 1,
 var Email: String = "")

The corresponding Java file is created and according to the Java Naming Conventions regading variables:

It should be a lowercase letter such as java, lang.

This means that doesn't matter if you choose the name of your fields to start either with an uppercase or a lowercase, all fiels are converted to start with a lowercase. That's why the properties in your database are present like that. Not only in Java/Kotlin is this convention available but also in other programming languages, so I recommend you to use this recommendation in your projects.

Regarding Kotlin, remember that at the end, everything is converted to byte code so it can work on JVM.

share|improve this answer

answered Mar 29 at 11:01

Alex MamoAlex Mamo

57.8k1111 gold badges4242 silver badges7878 bronze badges

data class User(var UsersName: String = "",
 var XP: Int = 0,
 var Level: Int = 1,
 var Email: String = "")

share|improve this answer

answered Mar 29 at 11:01

Alex MamoAlex Mamo

57.8k1111 gold badges4242 silver badges7878 bronze badges

answered Mar 29 at 11:01

Alex MamoAlex Mamo

57.8k1111 gold badges4242 silver badges7878 bronze badges

answered Mar 29 at 11:01

Alex MamoAlex Mamo

57.8k1111 gold badges4242 silver badges7878 bronze badges