Scala Playframework POST参数
我试图用PlayFrameworkScala(2.3)做一些非常简单的事情:创建一个路由帖子并获取帖子参数 路线定义Scala Playframework POST参数,scala,playframework,Scala,Playframework,我试图用PlayFrameworkScala(2.3)做一些非常简单的事情:创建一个路由帖子并获取帖子参数 路线定义 POST /ff/user controllers.Application.createUser 控制器 def createUser = Action.async { request => val user = request.queryString.get("user").flatMap(_.headOption)
POST /ff/user controllers.Application.createUser
控制器
def createUser = Action.async { request =>
val user = request.queryString.get("user").flatMap(_.headOption)
val email = request.queryString.get("email").flatMap(_.headOption)
val firstname = request.queryString.get("firstname").flatMap(_.headOption)
val lastname = request.queryString.get("lastname").flatMap(_.headOption)
Logger.debug("Create User")
Logger.debug(s"user=$user")
Logger.debug(s"email=$email")
Ok("Youpi")
}
当我向/ff/user发送请求时,日志显示:user=None,email=None。
我不明白他们为什么“没有”。怎么了
感谢您的帮助。当您像这样使用
POST
时,您可能希望查看请求
参数上的正文
字段,该字段将包含已发布的表单。您通常不会将查询字符串用于POST
请求(更多信息)。所以,这可能看起来像:
def createUser = Action.async { request =>
val user = request.body.asFormUrlEncoded.get.get("user").head
Future(Ok())
}
def createUser = Action.async(parse.urlFormEncoded) { request =>
//body is already treated as a Map[String, Seq[String]] because of the parameter passed to async
val user = request.body("user").head
Future(Ok())
}
您可能还希望使用提供解析正文的Action.async
重载。例如,它可能看起来像:
def createUser = Action.async { request =>
val user = request.body.asFormUrlEncoded.get.get("user").head
Future(Ok())
}
def createUser = Action.async(parse.urlFormEncoded) { request =>
//body is already treated as a Map[String, Seq[String]] because of the parameter passed to async
val user = request.body("user").head
Future(Ok())
}