我有类型为 Checkbox1
- Annot1
- Mouseup
- Action1
的域对象,它在案例类中定义如下:
Person
和相应的查询表如下:
case class Person(personName: String, personAge: Int, personId: Long = 0)
并且我有类型final case class PersonTable(tag: Tag) extends Table[Person](tag, "people") {
def personId = column[Long]("person_id", O.AutoInc, O.PrimaryKey)
def personName = column[String]("person_name")
def age = column[Int]("person_age")
def * = (personName, age, personId).mapTo[Person]
}
的域对象,它在案例类中定义如下:
Address
其对应的表类定义如下。人与地址具有一对多关系( 一个人可以有多个地址,但一个地址仅属于一个人 )
case class Address(houseNumber: Int, street: String
, state: String, ownerId: Long, id: Long = 0)
我想从基于final case class AddressTable(tag: Tag) extends Table[Address](tag, "addresses") {
def houseNumber = column[Int]("house_number")
def street = column[String]("street_name")
def state = column[String]("state_name")
def addressId = column[Long]("address_id", O.PrimaryKey, O.AutoInc)
def ownerId = column[Long]("owner_id")
def owner = foreignKey("owner_fk", ownerId, TableQuery[PersonTable])(_.personId, onDelete = ForeignKeyAction.Cascade)
def * = (houseNumber, street, state, ownerId, addressId).mapTo[Address]
}
的REST API中返回以下类型的JSON结果(地址应作为数组返回):
Akka-http
要获得此结果,我使用Slick编写了查询,如下所示:
{
"name": "shekhar",
"age": 30,
"id": 1234,
"addresses": [{
"house_number": 1,
"street": "water street",
"state": "foo bar",
"owner_id": 1234,
"address_id": 9874
},
{
"house_number": 99,
"street": "foo bar street",
"state": "foo disk state",
"owner_id": 1234,
"address_id": 007
}
]
}
我得到以下结果的结果(每个地址的人员详细信息):
val peopleQueries = TableQuery[PersonTable]
val addressQueries = TableQuery[AddressTable]
val query = peopleQueries.filter(_.personName === "Shekhar") joinLeft addressQueries on (_.personId === _.ownerId)
val futureResultData = db.run(query.result)
要将以上内容转换为预期的JSON格式,我可以等到数据库查询运行后再编写一些Scala代码以使其达到所需的格式,但这将阻止请求,直到此后处理完成为止(如果我对此有误)。
我的代码如下所示:
Vector(
(Person("shekhar",30,1234),
Some(Address(1,"Water Street","foo bar",1234,9874))
)
, (Person("shekhar",30,1234),
Some(Address(99,"foo bar street","foo disk state",1234,007))
)
)
我想知道是否有任何方法可以得到非随机/非阻塞方式的结果?
感谢Pedro,我找到了解决此问题的方法。
对我有用的解决方案如下:
futureResultData.onComplete {
case Success(data) => // code to bring data in desired format
}
答案 0 :(得分:2)
futureResultData.map {
_.transformInDesiredFormat // code to bring data in desired format
}
如果您发布的是可以转化的实际代码,我可以为您提供更多帮助
未来是单子,所以当您映射时,您将使用该未来中的东西。如果您拥有Future [Int],并且将其映射到未来中,那么您将与Int一起使用。
示例:
val aFuture = Future.successful(1)
aFuture.map(f=> f+ 1)