类型不匹配的 Scala Slick 查询

Question

我在 scala 中使用 slick 获取用户名和密码时遇到问题，基本上类似于

var query = "SELECT * FROM  \"default\".users as A " + "WHERE " + " A.username LIKE \'" + email + "\' " + "AND" + " A.password LIKE \'" + password + "\' ";

这是我的案例class 架构

  case class User(
                       id: Long = 0L,
                       username: String,
                       password: String,

                       author_id:Long,
                       created_on: DateTime,
                       updated_by: Long,
                       updated_on:Option[DateTime]
                     )

  class UserTable(tag:Tag) extends Table[User](tag,"user"){
    override def * = (id,username,password,author_id,created_on,updated_by,updated_on.?) <> (User.tupled,User.unapply)

    def id = column[Long]("id",O.PrimaryKey,O.AutoInc)

    def username = column[String]("username")

    def password = column[String]("password")

    def created_on = column[DateTime]("created_on")

    def updated_on = column[DateTime]("dateupdated")

    def author_id = column[Long]("author_id")

    def updated_by = column[Long]("updated_by")
  }

 lazy  val UserTable = TableQuery[UserTable]

下面是我的查询

val users = Main.UserTable.filter(_.username == email)
      .filter(_.password  == password).take(1)

Answer 1

filter(_.username == email)

您的意思可能是 === 而不是 ==。

这是用于构建查询表达式的 Slick DSL 的一部分（他们不能称之为 == 因为 Scala 的相等比较器无法被替换）。

Warning: Most operators mimic the plain Scala equivalents, but you have to use === instead of == for comparing two values for equality and =!= instead of != for inequality. This is necessary because these operators are already defined (with unsuitable types and semantics) on the base type Any, so they cannot be replaced by extension methods.

类型不匹配的 Scala Slick 查询

Type Mistmatch Scala Slick Query

database

postgresql

scala

slick