如何将主键作为Room Persistence lib的自动增量

Question

我正在创建一个Entity(Room Persistence lib)类Food,我想把它foodId作为自动增量.

@Entity
class Food(var foodName: String, var foodDesc: String, var protein: Double, var carbs: Double, var fat: Double)
{
    @PrimaryKey
    var foodId: Int = 0
    var calories: Double = 0.toDouble()
}

如何设置foodId自动增量字段？

Answer 1

您需要使用该autoGenerate属性

您的主键注释应如下所示:

@PrimaryKey(autoGenerate = true)

参考这里

Answer 2

您可以@PrimaryKey(autoGenerate = true)像这样添加:

@Entity
data class Food(
        var foodName: String, 
        var foodDesc: String, 
        var protein: Double, 
        var carbs: Double, 
        var fat: Double
){
    @PrimaryKey(autoGenerate = true)
    var foodId: Int = 0 // or foodId: Int? = null
    var calories: Double = 0.toDouble()
}

Answer 3

经过这么多答案后令人难以置信，但最后我的做法几乎没有什么不同。我不喜欢主键可以为空，我想将它作为第一个参数，并且还想在不定义它的情况下插入，而且它也不应该是 var。

@Entity(tableName = "employments")
data class Employment(
    @PrimaryKey(autoGenerate = true) val id: Long,
    @ColumnInfo(name = "code") val code: String,
    @ColumnInfo(name = "title") val name: String
){
    constructor(code: String, name: String) : this(0, code, name)
}

Answer 4

加 @PrimaryKey(autoGenerate = true)

@Entity
public class User {

    public User(int id, String name, String phone) {
        this.id = id;
        this.name = name;
        this.phone = phone;
    }


    @PrimaryKey(autoGenerate = true)
    private int id;

    @ColumnInfo(name = "full_name")
    private String name;

    @ColumnInfo(name = "phone")
    private String phone;

}

同时存储数据

 db.userDao().InsertAll(new User(0,sName,sPhone));

在创建对象时将id设为零(我的用例对象)

如果字段类型为long或int(或其TypeConverter将其转换为long或int),则Insert方法在插入项时将0视为未设置.

如果字段的类型是Integer或Long(Object)(或者其TypeConverter将其转换为Integer或Long),则Insert方法在插入项时将null视为未设置.

Answer 5

这对我有用：

@Entity(tableName = "note_table")
data class Note(
    @ColumnInfo(name="title") var title: String,
    @ColumnInfo(name="description") var description: String = "",
    @ColumnInfo(name="priority") var priority: Int,
    @PrimaryKey(autoGenerate = true) var id: Int = 0//last so that we don't have to pass an ID value or named arguments
)

请注意，在将实体插入 Room 之前，id 是最后一个以避免在创建实体时必须使用命名参数。将其添加到房间后，在更新实体时使用 id。

Answer 6

例如，如果您有一个users要存储的实体，带有字段(firstname, lastname , email)并且您想要自动生成的 id，则可以执行此操作。

@Entity(tableName = "users")
data class Users(
   @PrimaryKey(autoGenerate = true)
   val id: Long,
   val firstname: String,
   val lastname: String,
   val email: String
)

Room 然后将自动生成并自动增加该id字段。

Answer 7

@Entity(tableName = "user")
data class User(

@PrimaryKey(autoGenerate = true)  var id: Int?,
       var name: String,
       var dob: String,
       var address: String,
       var gender: String
)
{
    constructor():this(null,
        "","","","")
}

Answer 8

在下面的示例中，当您创建新用户时，请按构造函数中的方式传递参数。Room 将自动生成 id。所有用户对象 id 已在 id setter 中设置为 int 默认值，因此不要调用 setId

@Entity
public class User {

    @PrimaryKey(autoGenerate = true)
    private int id;

    @ColumnInfo(name = "full_name")
    private String name;

    @ColumnInfo(name = "phone")
    private String phone;

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

    public void setId(int id){
        this.id = id;
    }

}