Playframework：[RuntimeException：java.lang.reflect.InvocationTargetException]

我想創建一個簡單的登錄基於Zentask示例 - zentask - playframework，但是當我單擊登錄按鈕調用Application.authenticate操作時，它給運行時異常。我已經標有行 - 錯誤Playframework：[RuntimeException：java.lang.reflect.InvocationTargetException]

[RuntimeException: java.lang.reflect.InvocationTargetException]

Application.java

public class Application extends Controller { 

......... 

public static class Login 
{ 
    public String email; 
    public String password; 

    public String validate() 
    { 
     if (User.authenticate(email, password) == null) { 
      return "Invalid user or password"; 
     } 
     return null; 
    } 
} 

    public static Result authenticate() 
    { 
     Form<Login> loginForm = form(Login.class).bindFromRequest(); //--- error 
     if(loginForm.hasErrors()) { 
      return badRequest(login.render(loginForm)); 
     } else { 
      session("email", loginForm.get().email); 
      return redirect(
       routes.Application.index() 
      ); 
     } 
    } 
}

我理解它是與在登錄級的驗證功能，因爲當我刪除調用用戶。在驗證函數中驗證它的工作原理沒有錯誤。但我無法弄清楚。

用戶類是作爲 -

@Entity 
public class User extends Model 
{ 
    @Id 
    @Constraints.Required 
    @Formats.NonEmpty 
    public String userId; 

    @OneToOne(cascade=CascadeType.PERSIST) 
    AccountDetails accDetails; 


    public static Model.Finder<String,User> find = new Model.Finder<String,User>(String.class, User.class); 



    // Authenticate the user details 
    public static User authenticate(String email, String password) 
    { 
     String tempId = AccountDetails.authenticate(email, password).userId; 

     return find.ref(tempId); 
    } 

    .. . . . . . . 

}

和AccountDetails類 -

@Entity 
public class AccountDetails extends Model 
{ 
    @Id 
    String userId; 

    @Constraints.Required 
    String emailId; 

    @Constraints.Required 
    String password; 

    public static Model.Finder<String,AccountDetails> find = 
      new Model.Finder<String,AccountDetails>(String.class, AccountDetails.class); 


    public static AccountDetails authenticate(String email, String password) 
    {  
     return find.where() 
      .eq("email", email) 
      .eq("password", password) 
      .findUnique(); 
    } 

}

任何幫助，將不勝感激。

非常感謝。

來源

2013-10-02 ac-lap

沒有導入聲明，沒有異常堆棧跟蹤，碎片代碼。除了你已經處理了這樣的問題之外，回答這個問題是相當頑固的。 –

我必須承擔分配，但如果這是你的堆棧跟蹤是什麼樣子：

play.api.Application$$anon$1: Execution exception[[RuntimeException: java.lang.reflect.InvocationTargetException]] 
       at play.api.Application$class.handleError(Application.scala:293) ~[play_2.10.jar:2.2.3] 
       at play.api.DefaultApplication.handleError(Application.scala:399) [play_2.10.jar:2.2.3] 
       at play.core.server.netty.PlayDefaultUpstreamHandler$$anonfun$3$$anonfun$applyOrElse$3.apply(PlayDefaultUpstreamHandler.scala:264) [play_2.10.jar:2.2.3] 
       at play.core.server.netty.PlayDefaultUpstreamHandler$$anonfun$3$$anonfun$applyOrElse$3.apply(PlayDefaultUpstreamHandler.scala:264) [play_2.10.jar:2.2.3] 
       at scala.Option.map(Option.scala:145) [scala-library.jar:na] 
       at play.core.server.netty.PlayDefaultUpstreamHandler$$anonfun$3.applyOrElse(PlayDefaultUpstreamHandler.scala:264) [play_2.10.jar:2.2.3] 
     ... 
     Caused by: java.lang.NullPointerException: null 
       at models.User.authenticate(User.java:26) ~[na:na] 
       at controllers.Application$Login.validate(Application.java:50) ~[na:na] 
       at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method) ~[na:1.8.0_05] 
       at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62) ~[na:1.8.0_05] 
       at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43) ~[na:1.8.0_05] 
       at java.lang.reflect.Method.invoke(Method.java:483) ~[na:1.8.0_05]

那麼問題的根源實際上是在由NPE在User類暗示。如果輸入僞造證書，您的finder將找不到任何內容，並在AccountDetails.authenticate（）中返回null。

所以在下面你的方法不檢查空和嘗試，並獲取用戶標識，這將導致NPE：

public static User authenticate(String email, String password) { 
    String tempId = AccountDetails.authenticate(email, password).userId; 

    return find.ref(tempId); 
}

如果你只是檢查null正確，你將獲得所需的功能：

public static User authenticate(String email, String password) { 
    User user = null; 

    AccountDetails accountDetails = AccountDetails.authenticate(email, password); 
    if (accountDetails != null) { 
     user = find.ref(accountDetails.userId); 
    } 

    return user; 
}

來源

2014-06-24 09:05:47

Playframework：[RuntimeException：java.lang.reflect.InvocationTargetException]

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