forms - Scala Play 表单验证 : different forms for one case class - is it possible?

Question

我正在寻找一种方法来为一个案例类使用两种不同的形式。我试图用额外的构造函数来做到这一点，但失败了。查看代码片段:

case class LoginDetails(password: String, field3: Option[Int], field4: String)

case class User(username: String, loginDetails: LoginDetails) {
   def this(username: String, password: String, field3: Option[Int], field4: String) = this(username, LoginDetails(password, field3, field4))
// some logic inside
    }

val loginDetailsForm = Form(
 mapping(
   "password" -> text,
   "field3" -> optional(number),
   "field4" -> text
 )(LoginDetails.apply)(LoginDetails.unapply))

val oldForm = Form(
 mapping(
   "username" -> email,
   "password" -> text,
   "field3" -> optional(number),
   "field4" -> text
 )(User.apply)(User.unapply))

val newForm = Form(
     mapping(
       "username" -> email,
       "loginDetails" -> loginDetailsForm.mapping
     )(User.apply)(User.unapply))

我想做的是支持两个 API(新的和旧的)，但是如果我有同一个案例类的两个副本，我将不得不在其余的许多地方复制代码逻辑的代码。有办法吗？

当然，给定的代码不起作用，只是作为示例使用。

谢谢!

Answer 1

您可以尝试将方法 applyOld 和 unapplyOld 添加到伴随对象，如下所示:

case class User(username: String, loginDetails: LoginDetails)
object User {
  def applyOld(username: String, password: String, field3: Option[Int], field4: String) = new User(username, LoginDetails(password, field3, field4))
  def unapplyOld(u: User): Option[(String, String, Option[Int], String)] =
    Some((u.username, u.loginDetails.password,  u.loginDetails.field3,  u.loginDetails.field4))
}

你只需要方法 (String, String, Option[Int], String) => User 和 User => Option[(String, String, Option[Int], String) ]，你可以把它们放在任何你喜欢的地方。例如，您可以使用方法 apply 和 unapply 创建对象 UserOld。

用法:

val oldForm = Form(
 mapping(
   "username" -> email,
   "password" -> text,
   "field3" -> optional(number),
   "field4" -> text
 )(User.applyOld)(User.unapplyOld))

val newForm = Form(
     mapping(
       "username" -> email,
       "loginDetails" -> loginDetailsForm.mapping
     )(User.apply)(User.unapply))