如何使用 quill 在运行时定义 table 名称

Question

我是 quill 的新手，我试图在运行时定义 table，但我遇到了编译错误。有什么解决方法吗？或者它根本不可能使用 quill ？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 {

Answer 1

从 quill 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

如何使用 quill 在运行时定义 table 名称

How to define table name at runtime using quill

scala

quill.io