Cannot find setter for field - using Kotlin with R

问题:

I'm integrating with the Room persistence library. I have a data class in Kotlin like:

@Entity(tableName = "story")
data class Story (
        @PrimaryKey val id: Long,
        val by: String,
        val descendants: Int,
        val score: Int,
        val time: Long,
        val title: String,
        val type: String,
        val url: String
)

The @Entity and @PrimaryKey annotations are for the Room library. When I try to build, it is failing with error:

Error:Cannot find setter for field.
Error:Execution failed for task ':app:compileDebugJavaWithJavac'.
> Compilation failed; see the compiler error output for details.

I also tried providing a default constructor:

@Entity(tableName = "story")
data class Story (
        @PrimaryKey val id: Long,
        val by: String,
        val descendants: Int,
        val score: Int,
        val time: Long,
        val title: String,
        val type: String,
        val url: String
) {
    constructor() : this(0, "", 0, 0, 0, "", "", "")
}

But this doesn't work as well. A thing to note is that it works if I convert this Kotlin class into a Java class with getters and setters. Any help is appreciated!

回答1:

Since your fields are marked with val, they are effectively final and don't have setter fields.

Try switching out the val with var. You might also need to initialize the fields.

@Entity(tableName = "story")
data class Story (
        @PrimaryKey var id: Long? = null,
        var by: String = "",
        var descendants: Int = 0,
        var score: Int = 0,
        var time: Long = 0L,
        var title: String = "",
        var type: String = "",
        var url: String = ""
)

回答2:

Hey I don't know if everyone know or not, but you can not have column which is starting from is into Room. For example you can't have like this

   @Entity(tableName = "user")
   data class User (
        @PrimaryKey var id: Long? = null,
        var userName: String = "",
        var isConnectedToFB: Boolean = false,
)

回答3:

This is a bug and is fixed in Room 2.1.0-alpha01

https://developer.android.com/jetpack/docs/release-notes#october_8_2018

Bug Fixes

• Room will now properly use Kotlin’s primary constructor in data classes avoiding the need to declare the fields as vars. b/105769985

回答4:

There is an issue in room db library java code generation.

I was using optional field isFavorite. It gives me same error then I change my field name to favorite then compiled.

before var isFavorite: Int? = 0, after changing working fine var favorite: Int? = 0, Thanks

回答5:

According to https://stackoverflow.com/a/46753804/2914140 if you have an autogenerated primary key, you should write so:

@Entity(tableName = "story")
data class Story (
        val by: String,
        val descendants: Int,
        val score: Int,
        val time: Long,
        val title: String,
        val type: String,
        val url: String
)  {
    @PrimaryKey(autoGenerate = true)
    var id: Int = 0
}

Note that @PrimaryKey is written inside the class body and contains modifier var.

If you later want to update a row in a database with different parameters, use these lines:

val newStory = story.copy(by = "new author", title = "new title") // Cannot use "id" in object cloning
newStory.id = story.id
dao.update(newStory)

UPDATE

I still don't use AndroidX, and Room is 'android.arch.persistence.room:runtime:1.1.1'.

You can extend this class from Serializable. But if you want to extend it from Parcelable, you will get a warning (over id variable): Property would not be serialized inro a 'Parcel'. Add '@IgnoredOnParcel' annotation to remove this warning:

Then I moved an id from the body to the constructor. In Kotlin I use @Parcelize to create Parcelable classes:

@Parcelize
@Entity(tableName = "story")
data class Story (
    @PrimaryKey(autoGenerate = true)
    var id: Int = 0,

    val by: String,
    val descendants: Int,
    val score: Int,
    val time: Long,
    val title: String,
    val type: String,
    val url: String
) : Parcelable

回答6:

Had this error in Java.

You cannot have a column starting with is or is_ in Java.

Try renaming the column.

Another solution:

You either have to pass the field in the constructor and initialize it with the constructor argument, or create a setter for it.

Example:

public MyEntity(String name, ...) {
   this.name = name;
   ...
}

public void setName(String name) {
    this.name = name;
}

回答7:

Another cause of this may be the naming of the field. If you use any of the pre-defined keywords, you will get the same error. For instance, you can not name your column "is_active".

Reference: http://www.sqlite.org/lang_keywords.html

回答8:

Just an update if somebody comes across this thread in 2019, after spending hours digging online on why this should work, but it doesn't.

Using val works as expected if you are using the AndroidX version ( androidx.room:room-<any>:2.*) but it doesn't when using the old android.arch.persistence.room:<any>:1.1.1 and it seems that version 2.* wasn't released on this latter repo.

Edit: typos

回答9:

If you want the val immutability available for your entity, it is possible.

1. You should update to AndroidX room current version.
2. Check for the related issue here it is marked as Won't Fix
3. Now they have release a fix related to the issue with version 2.0.0-beta01
4. Now you can use immutable val with default value e.g:

@Entity("tbl_abc")
data class Abc(
    @PrimaryKey
    val id: Int = 0, 
    val isFavourite: Boolean = false
)

Previously, the above snippet will throw an error of Cannot find setter for field. Changing into var is a great workaround, but I prefer for the entity class to be immutable from outside invocation

回答10:

You can now start your field with is but you can't have a number next to the is like : is2FooSelected, you have to rename to isTwoFooSelected.