我试图将不可变的Java类转换为Kotlin,但失败了。
前提 条件:
global build.gradle:
classpath "org.jetbrains.kotlin:kotlin-gradle-plugin:$kotlinVersion"
Module build.gradle
apply plugin: 'kotlin-android'
apply plugin: 'kotlin-android-extensions'
apply plugin: 'kotlin-kapt'
...
kapt "android.arch.persistence.room:compiler:$roomVersion"
...
implementation "org.jetbrains.kotlin:kotlin-stdlib-jdk7:$kotlinVersion"
Java 会议室实体:
@Immutable
@Entity
public class User {
@PrimaryKey
@SerializedName("user_id")
@Expose
private final int userId;
@SerializedName("display_name")
@Expose
private final String userName;
@SerializedName("amount")
@Expose
private final String amount;
@SerializedName("has_access")
@Expose
private final String hasAccess;
public User(final int userId, final String userName,
final String amount, final String hasAccess) {
this.userId = userId;
this.userName = userName;
this.amount = amount;
this.hasAccess = hasAccess;
}
public int getUserId() {
return userId;
}
public String getUserName() {
return userName;
}
public String getAmount() {
return amount;
}
public String getHasAccess() {
return hasAccess;
}
}
转换为 Kotlin 的同一实体:
class User(@field:PrimaryKey
@field:SerializedName("user_id")
@field:Expose
val userId: Int,
@field:SerializedName("display_name")
@field:Expose
val userName: String,
@field:SerializedName("amount")
@field:Expose
val amount: String,
@field:SerializedName("has_access")
@field:Expose
val hasAccess: String)
Java 实体工作没有任何问题,但转换为 Kotlin 会导致下一个 buid 错误:
e: error: Entities and Pojos must have a usable public constructor. You can have an empty constructor or a constructor whose parameters match the fields (by name and type).
e: error: Cannot find setter for field.
摘要问题:如何正确使用不可变的 Kotlin 实体和房间持久性库?
更新:用户数据库和 dao 位于与用户模型模块分开的位置。似乎 Room 不适用于 Kotlin 为 val 属性添加的构造函数括号中的@NotNull、@NonNull、@Nullable。即使将它们添加到 java 构造函数中也会导致相同的错误。
而不是使用普通的java class
在kotlin中使用data class
,你会得到所有的getter,equals,hashCode方法的权利。
data class User(@PrimaryKey @ColumnInfo(name = "user_id") val userId: Long,
@ColumnInfo(name ="display_name") val userName: String = "",
@ColumnInfo(name ="amount") val amount: String = "",
@ColumnInfo(name ="has_access") val hasAccess: String = "")