Play for Scala 中的 JPA 查询抛出编译错误

Question

我正在尝试在 Play for Scala 中实现 JPA 查询。我从 here 中获取信息，但由于示例在 Java 中，所以我可能弄错了。这是代码：

class ManageBanks @Inject() (jpaApi: JPAApi) extends Controller {

   @Transactional
   def readMany = {
      val em = jpaApi.em

      jpaApi.withTransaction( (em: EntityManager) => {   //  <-- error in this line
          val query = em.createQuery("from BankHib order by name")
          val list = query.getResultList.asScala.toList.map(_.asInstanceOf[BankHib])
          list
      })
   }
}

编译时出现如下错误：

overloaded method value withTransaction with alternatives: [T](x: String, x: Boolean, x: java.util.function.Supplier[T])T (x: Runnable)Unit [T](x: java.util.function.Supplier[T])T [T](x: String, x: Boolean, x: java.util.function.Function[javax.persistence.EntityManager,T])T [T](x: String, x: java.util.function.Function[javax.persistence.EntityManager,T])T [T](x: java.util.function.Function[javax.persistence.EntityManager,T])T cannot be applied to (javax.persistence.EntityManager ⇒ List[admin.manage.BankHib])

这段代码有什么问题？如何使查询工作？

Answer 1

看看这是否有效

class ManageBanks @Inject() (jpaApi: JPAApi) extends Controller {

   @Transactional
   def readMany = {


      jpaApi.withTransaction( () => {   //  <-- error in this line
          val em = JPA.em() // or jpaApi.em
          val query = em.createQuery("from BankHib order by name")
          val list = query.getResultList.asScala.toList.map(_.asInstanceOf[BankHib])
          list
      })
   }
}

Answer 2

这是因为 play.db.jpa.JPAApi.withTransaction 具有以下签名：

withTransaction(java.util.function.Function<javax.persistence.EntityManager,T>)
withTransaction(java.lang.String, java.util.function.Function<javax.persistence.EntityManager,T>)
withTransaction(java.lang.String, boolean, java.util.function.Function<javax.persistence.EntityManager,T>)
withTransaction(java.util.function.Supplier<T>)
withTransaction(java.lang.Runnable)
最后 withTransaction(java.lang.String, boolean, java.util.function.Supplier<T>)。

但是您传递的是 (javax.persistence.EntityManager ⇒ List[admin.manage.BankHib]) 类型的 Scala 函数。因此，错误的类型和编译器抱怨说它无法找到替代方案。

正确的方法是使用 java.util.function.Function[EntityManager, List]:

class ManageBanks @Inject()(jpaApi: JPAApi) extends Controller {

  def readMany = {
    jpaApi.withTransaction(new java.util.function.Function[EntityManager, List[BankHib]] {
      override def apply(em: EntityManager): List[BankHib] = {
        val query = em.createQuery("from BankHib order by name")
        query.getResultList.asScala.map(_.asInstanceOf[BankHib]).toList
      }
    })
  }
}

此外，请注意，您不需要混合使用 JPAApi 和 JPA 来获得 EntityManager，因为 JPAApi 的方法已将其提供给给定的功能。

Play for Scala 中的 JPA 查询抛出编译错误

JPA query in Play for Scala throws compilation error

scala

playframework

playframework-2.0

playframework-2.5