scala - 如何在运行时使用鹅毛笔定义表名

我在羽毛笔上是新手，我试图在运行时定义表，但我收到了编译错误。有什么解决方法吗？还是使用羽毛笔是不可能的？
代码示例是:

case class ExampleCaseClass(id : String, version : String)
class Example (db: CassandraAsyncContext[SnakeCase] , table : String ) {
    import db._

    def selectByVarId = quote {
        (id: String, version: String) =>
            querySchema[ExampleCaseClass](table).filter(example => (example.id == id) && (example.version == version))
    }
}

和错误:

Error:(114, 36) Tree Error:(124, 25) Tree 'Example.this.db.querySchema[***.ExampleCaseClass](Example.this.table)' can't be parsed to 'Ast'
    def selectById = quote {

最佳答案

从鹅毛笔3. +使用dynamicQuerySchema[Person]("Person") dynamic-query-api

    import io.getquill.{Literal, MirrorSqlDialect, SqlMirrorContext}
    import java.time.LocalDate

    final case class PersonId(value: Int) extends AnyVal

    final case class Person(
      id: PersonId,
      firstName: String,
      lastName: String,
      birthDate: LocalDate)

    val context = new SqlMirrorContext(MirrorSqlDialect, Literal)

    val person = "Person"
    val person2 = "Person2"

    import context._

    val personContext = dynamicQuerySchema[Person](person)
    val person2Context = dynamicQuerySchema[Person](person2)
    val query = translate(personContext.filter(_.id == lift(PersonId(1))))
    val query2 = translate(person2Context.filter(_.id == lift(PersonId(1))))
    println(s"$query")
    println(s"$query2")

输出

SELECT v0.id, v0.firstName, v0.lastName, v0.birthDate, v0.addressId FROM Person v0 WHERE v0.id = 1
SELECT v0.id, v0.firstName, v0.lastName, v0.birthDate, v0.addressId FROM Person2 v0 WHERE v0.id = 1

关于scala - 如何在运行时使用鹅毛笔定义表名，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/44353397/