莫氏与房间——映射关系

Question

我有 Json，我想用 Moshi 映射并用 Room

存储

{
    "name": "My Group",
    "members": [
        {
            "id": "119075",
            "invitedUser": {
                "id": 97375,
                "email": "xxx@gmail.com"
            },
            "inviting_user": {
                "id": 323915,
                "email": "yyy@gmail.com"
            }
        },
        {
            "id": "395387",
            "invitedUser": {
                "id": 323915,
                "email": "aaa@gmail.com"
            },
            "inviting_user": {
                "id": 323915,
                "email": "bbb",
            }
        }
    ]
}

我准备好了我的模型

@Entity(tableName = "groups")
data class Group(
    @PrimaryKey
    val id: Long,
    val members: List<Member>
)

@Entity(tableName = "members")
data class Member(
    @PrimaryKey
    val id: Long,

    @Json(name = "invited_user")
    @ColumnInfo(name = "invited_user")
    val invitedUser: User,

    @Json(name = "inviting_user")
    @ColumnInfo(name = "inviting_user")
    val invitingUser: User
)

@Entity(tableName = "users")
data class User(
    @PrimaryKey
    val id: Int,

    val email: String
)

目前，我有 error: Cannot figure out how to save this field into database. 我读了这个 https://developer.android.com/training/data-storage/room/relationships。但是，如果我要像在文档中那样对关系建模，我不知道如何让 Moshi 映射关系？您找到该问题的最简单解决方案了吗？

Answer 1

我认为你有 2 个选择。

您将组和用户拆分到单独的表中并分别插入。
您使用 TypeConverters 将成员存储为组字段。

您的实施将取决于您的用例。

Answer 2

您使用 TypeConverters 将成员存储为组字段。

我相信这就是您需要的实现。

open class UserRequestConverter {

    private val moshi = Moshi.Builder().build()

    @TypeConverter
    fun fromJson(string: String): User? {
        if (TextUtils.isEmpty(string))
            return null

        val jsonAdapter = moshi.adapter(User::class.java)
        return jsonAdapter.fromJson(string)
    }

    @TypeConverter
    fun toJson(user: User): String {
        val jsonAdapter = moshi.adapter(User::class.java)
        return jsonAdapter.toJson(user)
    }
}



@Entity(tableName = "members")
data class Member(
    @PrimaryKey
    val id: Long,

    @Json(name = "invited_user")
    @ColumnInfo(name = "invited_user")
    @TypeConverters(UserRequestConverter::class)
    val invitedUser: User,

    @Json(name = "inviting_user")
    @ColumnInfo(name = "inviting_user")
    @TypeConverters(UserRequestConverter::class)
    val invitingUser: User
)

Answer 3

最后，我用TypeConverters

存储了它

private val membersType = Types.newParameterizedType(List::class.java, Member::class.java)
private val membersAdapter = moshi.adapter<List<Member>>(membersType)

@TypeConverter
fun stringToMembers(string: String): List<Member> {
    return membersAdapter.fromJson(string).orEmpty()
}

@TypeConverter
fun membersToString(members: List<Member>): String {
    return membersAdapter.toJson(members)
}

那是我的模特

@TypeConverters(Converters::class)
@Entity(tableName = "groups")
data class Group(
    @PrimaryKey
    val id: Long,
    val name: String
) {
    companion object {

        data class Member(
            val id: Long,
            val invitedUser: User,
            val invitingUser: User
        )

        data class User(
            val id: Long,
            val email: String
        )
    }
}

你觉得好看吗？可能更干净的方法是只拥有 ID 并将用户存储在其他地方，但我喜欢这个解决方案是如此简单。

莫氏与房间——映射关系

Moshi and room - mapping relationships

android

android-room

moshi

android-room-relation