我想在代码中创建以下配置。
mail{
signupUrl = "/test/ws/users/signup/"
signupFrom="Test<[email protected]>"
signupReply="Test<[email protected]>"
}
我尝试了以下两种方法,但我遇到了错误
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,需要一个Map
。
val mailConfig = Map("signupUrl" -> "/test/ws/users/signup/",
"signupFrom"->"Test<[email protected]>",
"signupReply"->"Test<[email protected]>")
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<[email protected]>",
"signupReply"->"Test<[email protected]>")
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<[email protected]>),(signupReply,Test<[email protected]>))
如果你只删除最后一个lin,你的第一个版本就可以了:
val mailConfig = Map("signupUrl" -> "/test/ws/users/signup/",
"signupFrom"->"Test<[email protected]>",
"signupReply"->"Test<[email protected]>")
val newConfig = Configuration.from(Map("mail"->mailConfig))
val newConfig: Configuration = Configuration.from(Map("mail"->mailConfig))
已经返回配置。
如果要从key -> value
对创建配置,则需要:
Seq
Seq
变换为varargs
(:_*
)。这里的例子如下:
val mailConfig = Seq("signupUrl" -> "/test/ws/users/signup/",
"signupFrom"->"Test<[email protected]>",
"signupReply"->"Test<[email protected]>")
val newConfiguration = Configuration(mailConfig: _*)