Play框架：不兼容的類型：java.util.List <models.Vehicle>不能轉換爲java.lang.String

Question

我試圖在Play框架2.5的簡單視圖，但我不斷收到此編譯時錯誤java.util.List<models.Vehicle> cannot be converted to java.lang.String

我已經嘗試過this answer

scala.collection.immutable.List<String> ls = JavaConverters.asScalaBufferConverter(scripts).asScala().toList(); 

但沒有.toList()功能，它無法識別。 這是我的代碼：

Aplication.java

package controllers; 

import models.Vehicle; 
import play.mvc.Controller; 
import play.mvc.Result; 
import scala.collection.JavaConverters; 

public class Application extends Controller { 

    public Result index() { 
    //  this does not work 
    //  scala.collection.immutable.List<Vehicle> ls = JavaConverters.asScalaBufferConverter(Vehicle.finder.all()).asScala().toList(); 
       return ok(views.html.index.render(Vehicle.finder.all())); //here is the error 
    } 

} 

index.scala.html

@(vehicles: java.util.List[Vehicle]) 

@main("Welcome to Play") { 

    <header> 
     <hgroup> 
      <h1>Dashboard</h1> 
      <h2>Vehicles</h2> 
     </hgroup> 
    </header> 

    <ul> 
    @for(vehicle <- vehicles) { 
     <li>@vehicle.getModel</li> 
    } 
    </ul> 
} 

相關的代碼在Vehicle.java

package models; 

import javax.persistence.*; 
import java.util.List; 
@Entity 
public class Vehicle extends BaseEntity { 

    public static Finder<Long, Vehicle> finder = new Finder<Long, Vehicle>(Vehicle.class){}; 
    @Column(nullable = false) 
    String model; 

    public String getModel() { 
     return model; 
    } 

    public void setModel(String model) { 
     this.model = model; 
    } 
} 

任何幫助將是非常讚賞。

Answer 1

好了，萬一有人認爲這在未來，嘗試

activator clean compile 

在項目文件夾

。

只做compile根本沒有幫助。