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

问题描述

我正在尝试基于Zentask示例创建一个简单的登录 - zentask - playframework ，但是当我单击调用Application.authenticate操作的登录按钮时，它会给出运行时异常。我已经标记了行 - 错误

I'm trying to create a simple login based on the Zentask sample --zentask - playframework, however when I click the login button which calls the Application.authenticate actions, it gives runtime exception. I have marked the line with -- error

[RuntimeException: java.lang.reflect.InvocationTargetException]

Application.java

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()
            );
        }
    }
}

我知道它有一些东西给使用Login Class中的validate函数，因为当我在validate函数中删除对User.authenticate的调用时，它可以正常工作。但是我无法弄明白。

I understand it has something to do with the validate function in Login Class, because when I remove the call to User.authenticate in the validate function it works without error. But I am unable to figure it out.

用户类是 -

@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类 -

and the AccountDetails Class -

@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();
    }

}

推荐答案

我必须承担很多，但如果这是你的堆栈跟踪的样子：

I have to assume a lot but if this is what your stacktrace looked like:

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]

然后问题的原因实际上是由您的User类中的NPE暗示的。
如果您输入伪造凭证，您的查找程序将找不到任何内容并在AccountDetails.authenticate（）中返回null。

then the cause of the problem is actually hinted at by the NPE in your User class. If you enter bogus credentials, your finder will not find anything and return null in AccountDetails.authenticate().

所以在下面的方法中，您不检查for null并尝试获取userId，这会导致NPE：

So in the method below you do not check for null and try and get the userId, which causes the NPE:

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

  return find.ref(tempId);
}

如果您只是正确检查null，您将获得所需的功能：

If you just check for null properly you will get the desired functionality:

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;
}

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