如何使用jax-rs子资源定位器处理持久化上下文（EntityManager）？ [英] How to handle persistence context (EntityManager) with jax-rs sub-resource locators?

问题描述

我在我的应用程序中使用jax-rs restful web服务和子资源定位器。但是，在将entityManager传递给子资源后，我无法在此子资源中保留任何新对象。

I am using jax-rs restful web service in my application with sub-resource locators. However after passing entityManager to sub-resource I cannot persist any new objects in this sub-resource.

然而，entityManager允许我查询数据。

The entityManager lets me however to query it for data.

这是我的主要资源：

@Path("/registrations")
@Stateless
public class RegistrationsResource {

    @Context
    private UriInfo context;

    @PersistenceContext(unitName="pctx")
    private EntityManager em;

    public RegistrationsResource() {
    }

    //POST method ommited

    @Path("{regKey}")
    public RegistrationResource getRegistrationResource(@PathParam("regKey")
    String regKey) {
        return RegistrationResource.getInstance(regKey, em);
    }

}

这是我的子资源：

public class RegistrationResource {

    private String regKey;
    private EntityManager em;

    private RegistrationResource(String regKey, EntityManager em) {
        this.regKey = regKey;
        this.em = em;
    }

    @Path("securityQuestion")
    @GET
    public String getQuestion() {
        return "iamahuman"+regKey;
    }

    @Path("securityQuestion")
    @POST
    public void postSecurityAnswer(String answer) {
        if(!answer.equals("iamahuman"+regKey)){
            throw new WebApplicationException(Status.BAD_REQUEST);
        }

        //Getting this information works properly
        List<RegistrationEntity> result = em.createNamedQuery("getRegistrationByKey")
            .setParameter("regKey", regKey).getResultList();

        switch(result.size()){
            case 0 :
                throw new WebApplicationException(Status.NOT_FOUND);
            case 1:
                break;
            default:
                throw new WebApplicationException(Status.INTERNAL_SERVER_ERROR);
            }

            RegistrationEntity reg = result.get(0);
            UserEntity newUser = new UserEntity();

            newUser.setHashedPassword(reg.getPwHash(), reg.getSalt());
            newUser.setUsername(reg.getUsername());
            newUser.setName(reg.getName());
            newUser.setSurname(reg.getSurname());

            //CRASHES HERE
            em.persist(newUser);
    }
}

正如您所看到的，它需要从数据库中注册对象，为注册创建新用户并尝试保留它。但是，em.persist（newUser）会抛出TransactionRequiredException。

As you can see it takes registration object from database, creates new user for registration and tries to persist it. However, em.persist(newUser) throws TransactionRequiredException.

我的问题是：我应该如何将EntityManager传递给子资源，以便它能够正确地保留新对象？

My question is: how should I pass EntityManager to sub-resource so it can properly persist new objects?

推荐答案

很抱歉再次挖掘这个，但我建议如下：

Sorry for digging this up again, but I suggest the following:

• 将子资源注释为@Stateless EJB


• 将@EJB注入成员字段放入父资源类中，如下所示：
  

• Annotate also the sub-resource as a @Stateless EJB
• Place @EJB injection member fields into the parent resource class, like so:

    @EJB private RegistrationResource registrationResource;

在getRegistrationResource（）中，不要调用子资源的构造函数，而是返回注入的EJB引用：

in "getRegistrationResource()", do not call the constructor of the sub-resource, but return the injected EJB reference:

public RegistrationResource getRegistrationResource() {
    return this.registrationResource;
}

但是，要使其工作，您不能将@PathParam作为构造函数参数传递。您必须通过@Context或其他@Path声明在子资源中单独访问它。

这使您能够以与父资源中完全相同的方式在子资源中注入EntityManager ，你不需要传递它。

For this to work, however, you cannot pass the "@PathParam" as a constructor parameter. You would have to access it seperately in the subresource via the "@Context" or another @Path declaration.
This enables you to inject the EntityManager in the sub resource in exactly the same way as in the parent resource, you don't need to pass it on.

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