Java ProvisionException:无法设置。游戏框架
当我将这行代码放入我的应用程序时,我无法再启动我的应用程序,我不知道为什么。Java ProvisionException:无法设置。游戏框架,java,playframework,Java,Playframework,当我将这行代码放入我的应用程序时,我无法再启动我的应用程序,我不知道为什么。 我用的是播放2.5 public class HomeController extends Controller { Form<User> userForm = Form.form(User.class); // get error if I put this line in public Hashtable<Integer, String> hmap = new Hash
我用的是播放2.5
public class HomeController extends Controller {
Form<User> userForm = Form.form(User.class); // get error if I put this line in
public Hashtable<Integer, String> hmap = new Hashtable<>();
公共类HomeController扩展控制器{
Form userForm=Form.Form(User.class);//如果我把这一行放进去,就会出现错误
公共哈希表hmap=新哈希表();
如果我在控制器中生成表单,就会得到错误
我得到的错误是:
ProvisionException: Unable to provision, see the following errors:
1) Error injecting constructor, java.lang.RuntimeException: There is no started application
at controllers.HomeController.<init>(HomeController.java:15)
while locating controllers.HomeController
for parameter 1 at router.Routes.<init>(Routes.scala:32)
while locating router.Routes
while locating play.api.inject.RoutesProvider
while locating play.api.routing.Router
for parameter 0 at play.api.http.JavaCompatibleHttpRequestHandler.<init>(HttpRequestHandler.scala:201)
while locating play.api.http.JavaCompatibleHttpRequestHandler
while locating play.api.http.HttpRequestHandler
for parameter 4 at play.api.DefaultApplication.<init>(Application.scala:221)
at play.api.DefaultApplication.class(Application.scala:221)
while locating play.api.DefaultApplication
while locating play.api.Application
1 error
ProvisionException:无法设置,请查看以下错误:
1) 注入构造函数java.lang.RuntimeException时出错:没有启动的应用程序
at controllers.HomeController.(HomeController.java:15)
查找控制器时。HomeController
用于router.Routes处的参数1。(Routes.scala:32)
定位路由器时。路由
查找play.api.inject.routeProvider时
查找play.api.routing.Router时
用于play.api.http.JavaCompatibleHttpRequestHandler处的参数0。(HttpRequestHandler.scala:201)
查找play.api.http.JavaCompatibleHttpRequestHandler时
查找play.api.http.HttpRequestHandler时
用于play.api.DefaultApplication中的参数4。(Application.scala:221)
at play.api.DefaultApplication.class(Application.scala:221)
查找play.api.DefaultApplication时
查找play.api.Application时
1错误
这是什么原因?有什么想法吗
编辑:
HomeController.java
package controllers;
import play.mvc.*;
import java.util.*;
import models.User;
import play.data.Form;
public class HomeController extends Controller {
Form<User> filledForm = Form.form(User.class);
public Hashtable<Integer, String> hmap = new Hashtable<>();
public Result index() {
return ok(views.html.index.render());
}
public Result saveData() {
User user = filledForm.bindFromRequest().get();
if (filledForm.hasErrors()) {
// process
} else {
// contactForm.get().firstName should be filled with the correct data
System.out.println("Nothing to show here.");
}
System.out.println(user.name);
System.out.println(user.tableNr);
return ok(views.html.index.render());
}
包控制器;
导入play.mvc.*;
导入java.util.*;
导入模型。用户;
导入play.data.Form;
公共类HomeController扩展控制器{
Form filledForm=Form.Form(User.class);
公共哈希表hmap=新哈希表();
公开结果索引(){
返回ok(views.html.index.render());
}
公共结果saveData(){
User=filledForm.bindFromRequest().get();
if(filledForm.hasErrors()){
//过程
}否则{
//contactForm.get().firstName应填写正确的数据
System.out.println(“此处无需显示”);
}
System.out.println(用户名);
System.out.println(user.tableNr);
返回ok(views.html.index.render());
}
您是如何运行的?完整代码。您指的是哪种代码?控制器类?