調用靜態方法，並把型號爲控制器

Question

我只是想給我的形式傳遞給我的控制器，我不管收到此錯誤是我嘗試：

render(play.api.data.Form<models.Service>) in 'null' cannot be applied to (play.data.Form<models.Service) 

錯誤行：

return ok(info.render(sServiceForm)); 

Info.scale.html - 查看

@(serviceForm : Form[Service]) 
@import helper._ 

@main("Service info") { 
    <h1>Service Information</h1> 
    @helper.form(action = routes.Services.save()) { 
    <fieldset> 
     <legend>Service</legend> 
     @helper.inputText(serviceForm.field("code"), '_label -> "Code") 
     @helper.inputText(serviceForm.field("description"), '_label -> "Description") 
     @helper.inputText(serviceForm.field("description"), '_label -> "Description") 
    </fieldset> 
    <input type="submit" value="Save" /> 
    } 
} 

Service.java - 型號

package models; 

import com.avaje.ebean.Model; 

import javax.persistence.Entity; 
import javax.persistence.Id; 

/** 
* Created by James on 3/4/2016. 
*/ 
// Telling play framework that this is a class thats going to map as a model to save service records 
@Entity 
public class Service extends Model { 
    // Internal ID to reference a certain activity 
    @Id 
    public String code; 
    public String description; 
} 

Services.java - 控制器

package controllers; 

import models.Service; 
import play.mvc.Controller; 
import play.mvc.Result; 
import play.data.Form; 
import views.html.services.info; 

/** 
* Created by James on 3/4/2016. 
*/ 
public class Services extends Controller { 
    // Creating static class variable, calling static method and passing our model class. 
    //private static final Form<Service> sServiceForm = Form.form(Service.class); 
    private static final Form<Service> sServiceForm = play.data.Form.form(Service.class); 
    public Result list() { 
     return TODO; 
    } 

    public Result addService() { 
     return ok(info.render(sServiceForm)); 
    } 

    public Result save() 
    { 
     return TODO; 
    } 
} 

如果我註釋掉：

private static final Form<Service> sServiceForm = Form.form(Service.class); 

而改變我的addService到return TODO;網站編譯良好，我可以遍歷很好。即使我仍然返回TODO，該行會破壞網站：

private static final Form<Service> sServiceForm = Form.form(Service.class); 

Answer 1

如果您希望它使用FormFactory工作。 您可以按照此代碼。 gitter的人幫助了我。所以，他的信用實際上。

這裏是我的代碼：

Services.java

package controllers; 

import models.Service; 
import play.data.Form; 
import play.data.FormFactory; 
import play.mvc.Controller; 
import play.mvc.Result; 
import views.html.services.info; 

import javax.inject.Inject; 

public class Services extends Controller { 
    private final Form<Service> serviceForm; 

    @Inject 
    public Services(FormFactory formFactory) { 
     this.serviceForm = formFactory.form(Service.class); 
    } 

    public Result list() { 
     return TODO; 
    } 

    public Result addService() { 
     return ok(info.render(serviceForm)); 
    } 

    public Result save() { 
     return TODO; 
    } 
} 

和info.scala.html

@(serviceForm : play.data.Form[Service]) 
@import helper._ 
@main("Service info"){ 

    <h1>Service Information</h1> 

    @form(action = routes.Services.save()) { 

    <fieldset> 

     <legend>Service</legend> 
     @inputText(serviceForm("code"), '_label -> "Code") 
     @inputText(serviceForm("description"), '_label -> "Description") 
     @inputText(serviceForm("description"), '_label -> "Description") 

    </fieldset> 

    <input type="submit" value="Save"/> 

    } 

}