无法创建配置对象

Question

我想在代码中创建以下配置。

mail{
  signupUrl = "/test/ws/users/signup/"
  signupFrom="Test<mailrobot@test.com>"
  signupReply="Test<noreply@test.comm>"

}

我尝试了以下两种方法，但出现错误

尝试 1. API (https://www.playframework.com/documentation/2.6.x/api/scala/index.html#play.api.Configuration$) 有一个 from 方法，该方法采用 Map.

val mailConfig = Map("signupUrl" -> "/test/ws/users/signup/",
        "signupFrom"->"Test<mailrobot@test.com>",
        "signupReply"->"Test<noreply@test.comm>")
      val newConfig = Configuration.from(Map("mail"->mailConfig))
      val newConfiguration = Configuration(newConfig)

错误

Error:(575, 30) overloaded method value apply with alternatives:
  (underlying: com.typesafe.config.Config)play.api.Configuration <and>
  (data: (String, Any)*)play.api.Configuration
 cannot be applied to (play.api.Configuration)
      val newConfiguration = Configuration(newConfig)

尝试 2 - API (https://www.playframework.com/documentation/2.6.x/api/scala/index.html#play.api.Configuration$) 有 apply 方法需要 (String,any*)

 val mailConfig = ("signupUrl" -> "/test/ws/users/signup/",
        "signupFrom"->"Test<mailrobot@test.com>",
        "signupReply"->"Test<noreply@test.comm>")
      val newConfig = ("mail"->mailConfig)
      val newConfiguration = Configuration(newConfig)

错误

bug in method caller: not valid to create ConfigValue from: ((signupUrl,/test/ws/users/signup/),(signupFrom,Test<mailrobot@test.com>),(signupReply,Test<noreply@test.comm>))

Answer 1

如果你只删除最后一个 lin，你的第一个版本就可以工作，比如：

val mailConfig = Map("signupUrl" -> "/test/ws/users/signup/",
        "signupFrom"->"Test<mailrobot@test.com>",
        "signupReply"->"Test<noreply@test.comm>")
      val newConfig = Configuration.from(Map("mail"->mailConfig))

val newConfig: Configuration = Configuration.from(Map("mail"->mailConfig))

已经returns一个配置。

如果您想从 key -> value 对创建配置，您需要：

将这些对放在 Seq
将 Seq 转换为 varargs (:_*)。

举个例子：

val mailConfig = Seq("signupUrl" -> "/test/ws/users/signup/",
      "signupFrom"->"Test<mailrobot@test.com>",
      "signupReply"->"Test<noreply@test.comm>")
    val newConfiguration = Configuration(mailConfig: _*)

无法创建配置对象

unable to create Configuration object

scala

playframework-2.6