Question

我使用的是Play 2.5和ReactiveMongo。这是一个简单的问题，但我无法弄清楚。我有一个控制器从视图中接收username和password，我想将此请求转换为类型UserModel的模型，然后将此模型转换为json并将其写入MongoDB

我的控制器：

class RegisterController @Inject() (val reactiveMongoApi: ReactiveMongoApi) 
extends Controller with MongoController with ReactiveMongoComponents {

  def registerPost = Action.async { request =>

  implicit val accountWrites = new Writes[UserModel] {
  def writes(account: UserModel) = Json.obj(
    "username" -> account.username,
    "password" -> account.password
  )
} //needs this for some reason?

val future = collection.insert(request.body) //trying to insert to mongo
future.map(_ => Ok)

我的模特：

case class UserModel(username: String, password: String) {}

object UserModel {

  implicit val format = Json.format[UserModel] //needs this for some reason?

  val userModel = Form(
mapping(
  "username" -> nonEmptyText,
  "password" -> nonEmptyText
)(UserModel.apply)(UserModel.unapply))
}

我的观点：

@(userForm: Form[UserModel])(implicit messages: Messages)

<h1>Register</h1>

@helper.form(action = routes.RegisterController.registerPost()) {
  @helper.inputText(userForm("username"))
  @helper.inputText(userForm("password"))

  <button type="submit" name="action" value="register">Register</button>

}

Answer 1

这些是您需要做的关键事项：

=＆GT;控制器方法Bind the form request中的registerPost。为此，您还需要form mapping设置。此映射将帮助您使用方法内的表单数据生成UserModel对象。要做到这一切，请参阅此处ScalaForms

=＆GT;你这里不需要writes。您可以使用format同时执行json writes and reads。

=＆GT;现在将UserModel转换为JsValue：

//usermodel will be generated by binding the form data
implicit val format = Json.format[UserModel]
val userModelJson = format.writes(usermodel)

=＆GT;然后，您只需将存储库称为：

collection.insert(userModelJson)

Scala Play将请求转换为JSON并将其写入MongoDB

1 个答案: