小编OZK*_*ZKA的帖子

我不明白我怎么能改写用于jetty 6的jetty 6的代码:

import org.mortbay.jetty.*;
import org.mortbay.jetty.nio.SelectChannelConnector;
import org.mortbay.jetty.webapp.WebAppContext;

public class ApplLauncher {
    public static void main(String[] args) {
        Server server = new Server();

        Connector connector = new SelectChannelConnector();
        connector.setPort(8080);
        server.addConnector(connector);

        WebAppContext root = new WebAppContext("C:\\Users\\OZKA\\IdeaProjects\\projectname\\projectname\\web", "/");
        root.setWelcomeFiles(new String[]{"index.html"});
        //root.addServlet(new ServletHolder(new TestServlet()), "/test");

        server.setHandlers(new Handler[]{root});

        try {
            server.start();
        } catch (Exception e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }

    }

}

上面的代码工作正常,并响应web.xml中映射的web文件夹和servlet中的静态内容.这是我尝试使用嵌入式码头9:

import org.eclipse.jetty.server.Server;
import org.eclipse.jetty.webapp.WebAppContext;
import org.eclipse.jetty.server.handler.ResourceHandler;
import org.eclipse.jetty.server.handler.HandlerList;
import org.eclipse.jetty.server.Handler;


public class ApplLauncher {
    public static void …

是否有一种方法和/或库可以自动从Json创建Kotlin Data类,就像在Scala Json.Spray中一样？

像这样的东西:

data class User(id: Int, name: String)

class DataClassFactory(val json: String) {
   fun getUser(): User {
      //some reflection
      return User(10, "Kirill")
  }
}

fun main(args: Array<String>): Unit {
  val json = "{id: 10, name: Kirill}"
  val usr = DataClassFactory(json).getUser()
  println(usr)
}

我有一个错误：

spray.json.ProductFormats$class.productElement2Field NullPointerException

这是我的 json 反序列化代码：

object DomainJsonProtocol extends DefaultJsonProtocol {
implicit val loginInfoFormat = jsonFormat(LoginInfo, "userid", "email", "password", "rememberme")
implicit val requestStatusFormat = jsonFormat(RequestStatus, "status", "message")
implicit val requestHolderFormat = jsonFormat(RequestHolder, "requestStatus", "loginInfo")
}

case class RequestHolder(requestStatus : RequestStatus, loginInfo: LoginInfo) {
  def this(requestStatus : RequestStatus) = this(requestStatus, null)
}
case class LoginInfo(userid: Int, email: String, password: String, rememberme: Boolean)
case class RequestStatus(status : Int, message: String)

val requestHolder = content.asJson.convertTo[RequestHolder] //The error is hereHere is 

我想这可能是因为类 RequestHolder 中的构造函数重载。 …

我有一个json对象"{"id":1,"name":"OZKA","birthDate":"1981-02-08T20:00:00.000Z","monthRevenue":1000.75,"developer":true}"和代码:

println(request.getParameter("content"))//{"id":1,"name":"OZKA","birthDate":"1981-02-08T20:00:00.000Z","monthRevenue":1000.75,"developer":true}
val result = scala.util.parsing.json.JSON.parseFull(request.getParameter("content"))
result match {
  case Some(e) => { println(e); //output: Map(name -> OZKA, monthRevenue -> 1000.75, developer -> true, birthDate -> 1981-02-08T20:00:00.000Z, id -> 1.0)
    e.foreach((key: Any, value: Any) => {println(key + ":" + value)})
  }
  case None => println("Failed.")
}

,当我尝试调用map或foreach函数时,编译器抛出一个错误"值foreach不是Any的成员".任何人都可以建议我一个方法,我如何解析这个json字符串并将其转换为Scala类型