使用 Jooq 创建通用存储库

Question

我们正在使用 JOOQ 来映射现有数据库。此数据库中的每个表（第 1 条）共享一组相似的列：

resource_id:   Int (FK)
id:            Int (PK)
identifier:    UUID
date_created:  DateTime
date_modified: DateTime

数据库中有一个通用规则，所有操作看起来像这样：

select   * 
from     table t
join     resource r on r.id = t.resource_id
where    r.is_archived = false 

使用 JOOQs 生成功能，我可以成功创建一个实现基本查询操作的存储库类：

@Repository
class UserRepository(val dsl:DSLContext): IRepository {
    val record = Tables.AUTHENTICATION_USER!!   //generated by jooq
    val resource = Tables.RESOURCE!!            //generated by jooq
    val pojo = AuthenticationUser::class.java   //generated by jooq

    // this is the general query logic that should be applied to all the things
    override fun selectQuery(): SelectConditionStep<Record> {
        return dsl
            .select(record.asterisk())
            .from(record)
            .join(resource).on(resource.ID.eq(record.RESOURCE_ID))
            .where(resource.IS_ARCHIVED.eq(false))
    }

    override fun asyncCount(): Mono<Int> = Mono.fromCallable {
        selectQuery()
            .count()
    }

    fun asyncGetPage(from: Int,
                     size: Int,
                     orderBy: SortField<*> = record.DATE_CREATED.asc(),
    ) = Mono.fromCallable {
        selectQuery()
            .orderBy(orderBy)
            .limit(size)
            .offset(from)
            .fetchInto(pojo)
            .let { Collections.unmodifiableList(it) }
    }

现在，这一切都很好，但我无法将其转换为可以实现的通用实现

• 接受table 类型T
• 可以识别表上的这 5 个常见字段/属性，因此我们可以使查询通用化。

我可以让它工作：

class GeneralRepository<T: Table<R>, R: Record>(val table: T, val record: R, val dsl:DSLContext) {
    fun asyncGetAll(): Mono<List<R>> = Mono.fromCallable {
        dsl
            .select()
            .from(table)
            .fetchInto(record::class.java)
    }
}

// manually construct it ignoring Springs DI for development
val resourceRepository = GeneralRepository(
    Tables.RESOURCE,
    ResourceRecord(),
    JooqConfiguration().dsl(),
)

当我尝试对table T 做任何事情时，问题就来了。由于T 是通用表类型，编译器无法知道id、identifier、resource_id、date_created 等字段，这意味着这样的查询无法编译：

    fun asyncGetAllNonArchived() = 
        dsl
            .select()
            .from(table)
            .where(table.DATE_CREATED > ...)

问题： 有没有办法告诉编译器或以公共字段可用的方式声明泛型。

---

编辑：感谢以下 Lukas Eder 的建议。通过调整该解决方案，我们能够创建一个通用存储库。

首先，将生成器策略更改为 Kotlin 让 Auditable 接口正确运行 :)

其次，我们无法使用泛型类来实现 Table 类型的 Record 类型，因此您必须稍微扩展签名。

interface Auditable {
    fun RESOURCE_ID(): TableField<Record, Int> =
        (this as Table<Record>).field("resource_id") as TableField<Record, Int>

    fun ID(): TableField<Record, Int> =
        (this as Table<Record>).field("id") as TableField<Record, Int>

    fun IDENTIFIER(): TableField<Record, UUID> =
        (this as Table<Record>).field("identifier") as TableField<Record, UUID>

    fun DATE_CREATED(): TableField<Record, LocalDateTime> =
        (this as Table<Record>).field("date_created") as TableField<Record, LocalDateTime>

    fun DATE_MODIFIED(): TableField<Record, LocalDateTime> =
        (this as Table<Record>).field("date_modified") as TableField<Record, LocalDateTime>
}


open class GeneralRepository<R, T>(val table: T) where R: Record, T: Table<R>, T: Auditable {
    val resource = Resource.RESOURCE

    fun selectQuery() =
        dsl
            .select(table.asterisk())
            .from(table)
            .join(resource).on(resource.ID.eq(table.RESOURCE_ID()))
            .where(resource.IS_ARCHIVED.eq(false))

然后您可以轻松实现通用扩展：

class UserRepo: GeneralRepository<UserRecord, User>(table = User.USER) {

    fun getEmail(email: String) = Mono.fromCallable {
        selectQuery().and(table.EMAIL.eq(email))
    }
}

Answer 1

您可以生成一个接口来提供对这些列的访问：

interface Auditable {
  fun RESOURCE_ID(): TableField<Record, Int?> = 
   (this as Table<Record>).field("resource_id") as TableField<Record, Int?>
  fun ID(): TableField<Record, Int?> = 
   (this as Table<Record>).field("id") as TableField<Record, Int?>
  fun identifier(): TableField<Record, UUID?> = 
   (this as Table<Record>).field("identifier") as TableField<Record, UUID?>
  fun DATE_CREATED(): TableField<Record, LocalDateTime?> = 
   (this as Table<Record>).field("date_created") as TableField<Record, LocalDateTime?>
  fun DATE_MODIFIED(): TableField<Record, LocalDateTime?> = 
   (this as Table<Record>).field("date_modified") as TableField<Record, LocalDateTime?>
}

然后使用generator strategy将接口附加到所有生成的表：

<strategy>
    <matchers>
        <tables>
            <table>
                <expression>T_.*</expression>
                <tableImplements>com.example.Auditable</tableImplements>
            </table>
        </tables>
    </matchers>
</strategy>

目前无法在接口中声明属性，因为无法检测是否需要override 修饰符，请参阅https://github.com/jOOQ/jOOQ/issues/10776。

但是现在这可能已经足够了。然后，您可以创建如下一般条件：

fun <T> dateCreatedCondition(table: T, dt: LocalDateTime): Condition
where T : Table<*>, T : Auditable = table.DATE_CREATED().gt(dt)