Question

添加以下代码后：

@GET
@Produces({"application/xml", "application/json"})
public WatchList find(@PathParam("id") Integer id) {
    WatchList results = em.createNamedQuery("WatchList.findById", WatchList.class)
            .setParameter("id", id)
            .getSingleResult();
    return results;
}

@GET
@Path("{id}")
@Produces(MediaType.TEXT_HTML)
public void findById(
        @Context final HttpServletRequest request,
        @Context final HttpServletResponse response) throws ServletException, IOException {
    RequestDispatcher dispatcher = request.getRequestDispatcher("/WEB-INF/jsp/AMLManagement.jsp");
    request.setAttribute("items", find("id"));
    dispatcher.forward(request, response);
}

我收到错误Severe: WebModule[/MavenProjectTest]StandardWrapper.Throwable org.glassfish.jersey.server.model.ModelValidationException: Validation of the application resource model has failed during application initialization.

而且：Severe: Exception while loading the app : java.lang.IllegalStateException: ContainerBase.addChild: start: org.apache.catalina.LifecycleException: org.apache.catalina.LifecycleException: org.glassfish.jersey.server.model.ModelValidationException: Validation of the application resource model has failed during application initialization.

有谁知道如何解决这个问题？我在上面的代码中做了什么错误导致了这个错误吗？

Answer 1

好的，我必须添加一个路径，我传递的id不应该是一个字符串，我必须在findId中添加@Pathparam，我解决了这个问题：

@GET
@Path("{id}")
@Produces({"application/xml", "application/json"})
public WatchList find(@PathParam("id") Integer id) {
    WatchList results = em.createNamedQuery("WatchList.findById", WatchList.class)
            .setParameter("id", id)
            .getSingleResult();
    return results;
}

@GET
@Path("{id}")
@Produces(MediaType.TEXT_HTML)
public void findId(@PathParam("id") Integer id,
        @Context final HttpServletRequest request,
        @Context final HttpServletResponse response) throws ServletException, IOException {
    RequestDispatcher dispatcher = request.getRequestDispatcher("/WEB-INF/jsp/AMLManagement.jsp");
    request.setAttribute("items", find(id));
    dispatcher.forward(request, response);
}

严重：StandardWrapper.Throwable在jax-rs maven项目中

1 个答案: