Re: [Wicket-user] RuntimeException not being caught despite configuration

[Jaime De La Jara]

I changed the init parameter "configuration" to deployment but the problem persists. If I change the following url :

http://localhost:8080/phonebook/app?wicket:interface=:1:1:

to

http://localhost:8080/phonebook/app?wicket:interface=:1:X:

I still get :

java.lang.NumberFormatException: For input string: "X"
 java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
 java.lang.Integer.parseInt(Integer.java:447)
 java.lang.Integer.parseInt(Integer.java:497)
 wicket.protocol.http.request.WebRequestCodingStrategy.addInterfaceParameters(WebRequestCodingStrategy.java:424)
 wicket.protocol.http.request.WebRequestCodingStrategy.decode(WebRequestCodingStrategy.java:157)
 wicket.Request.getRequestParameters(Request.java:161)
 wicket.Page.init(Page.java:1149)
 wicket.Page.<init>(Page.java:194)
 wicket.markup.html.WebPage.<init>(WebPage.java:119)

 wicket.contrib.phonebook.web.page.BasePage.<init>(BasePage.java:53)
 wicket.contrib.phonebook.web.page.ErrorPage.<init>(ErrorPage.java:38)
 wicket.contrib.phonebook.web.PhonebookApplication$2$1.onRuntimeException(PhonebookApplication.java:250)
 wicket.request.compound.DefaultExceptionResponseStrategy.onRuntimeException(DefaultExceptionResponseStrategy.java:125)
 wicket.request.compound.DefaultExceptionResponseStrategy.respond(DefaultExceptionResponseStrategy.java:64)
 wicket.request.compound.AbstractCompoundRequestCycleProcessor.respond(AbstractCompoundRequestCycleProcessor.java:76)
 wicket.RequestCycle.step(RequestCycle.java:1006)
 wicket.RequestCycle.steps(RequestCycle.java:1040)
 wicket.RequestCycle.request(RequestCycle.java:454)
 wicket.protocol.http.WicketServlet.doGet(WicketServlet.java:216)
 javax.servlet.http.HttpServlet.service(HttpServlet.java:689)
 javax.servlet.http.HttpServlet.service(HttpServlet.java:802)

 org.springframework.orm.jdo.support.OpenPersistenceManagerInViewFilter.doFilterInternal(OpenPersistenceManagerInViewFilter.java:106)
 org.springframework.web.filter.OncePerRequestFilter.doFilter(OncePerRequestFilter.java:77)

The config I'm using is :
<context-param>
    <param-name>configuration</param-name>
    <param-value>deployment</param-value>
  </context-param>

Surely I'm confusing some parameter value or something.

Jaime.


>>i think you have to be in deployment mode

>>-Igor


>>>>On 10/11/06, Jaime De La Jara <[EMAIL PROTECTED] > wrote:Hi, changing the >>>>url information and reloading causes a RuntimeException stack trace to be >>>>shown instead of a custom ErrorPage that I have configured in y app :

>>>>This is the configuration being used :

>>>>getApplicationSettings().setInternalErrorPage(ErrorPage.class);
    >>>>getExceptionSettings().setUnexpectedExceptionDisplay(IExceptionSettings.SHOW_INTERNAL_ERROR_PAGE >>>>);


>>>>and

>>>> protected IRequestCycleFactory getDefaultRequestCycleFactory()
>>>>  {
>>>>    return new IRequestCycleFactory()
>>>>    {
>>>>      private static final long serialVersionUID = 1L;
>>>>
>>>>      public RequestCycle newRequestCycle(Session session,
>>>>                                          Request request, Response response)
>>>>      {
>>>>        return new WebRequestCycle((WebSession) session,
>>>>                                   (WebRequest) request, (WebResponse) response)
>>>>        {
>>>>           @Override
>>>>           public wicket.Page onRuntimeException(wicket.Page page,   
>>>>                                                                       RuntimeException e)
>>>>           {
>>>>             System.out.println("Page : " + page);
>>>>             return new ErrorPage(e);
>>>>            }
>>>>        };
>>>>      }
>>>>    };
>>>>   }

>>>>and ErrorPage constructor is :

>>>>  public ErrorPage(RuntimeException e)
>>>>  {
>>>>    String errorMsg = "Se ha producido un error inesperado";
>>>>    System.out.println("Exception : " + getRootException(e));
>>>>    if (UnAuthorizedException.class.isInstance (getRootException(e)))
>>>>      errorMsg = "Ud. no esta autorizado a realizar esta accion";
>>>>    add(new Label("error", errorMsg));
>>>>    add(new Link("homeLink")
>>>>    {
>>>>      public void onClick()
>>>>      {
>>>>        setResponsePage(new ListContactsPage());
>>>>      }
>>>>    });
>>>>    }

>>>>However the following trace is shown when I tweak the Url (as a user could >>>>do) and press the reload button :

>>>>java.lang.NumberFormatException: For input string: "users"

>>>> java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
>>>> java.lang.Integer.parseInt(Integer.java:447)
>>>> java.lang.Integer.parseInt(Integer.java:497)
>>>> wicket.protocol.http.request.WebRequestCodingStrategy.addInterfaceParameters
>>>>(WebRequestCodingStrategy.java:424)
>>>> wicket.protocol.http.request.WebRequestCodingStrategy.decode(WebRequestCodingStrategy.java:157)
>>>> wicket.Request.getRequestParameters(Request.java:161)
>>>> wicket.Page.init(Page.java
>>>>:1149)
>>>> wicket.Page.<init>(Page.java:194)
>>>> wicket.markup.html.WebPage.<init>(WebPage.java:119)
>>>> wicket.contrib.phonebook.web.page.BasePage.<init>(BasePage.java:53)
>>>> wicket.contrib.phonebook.web.page.ErrorPage
>>>>.<init>(ErrorPage.java:37)
>>>>
 wicket.contrib.phonebook.web.PhonebookApplication$2$1.onRuntimeException(PhonebookApplication.java:249)
>>>> wicket.request.compound.DefaultExceptionResponseStrategy.onRuntimeException(
>>>>DefaultExceptionResponseStrategy.java:125)

>>>> wicket.request.compound.DefaultExceptionResponseStrategy.respond(DefaultExceptionResponseStrategy.java:64)
>>>> wicket.request.compound.AbstractCompoundRequestCycleProcessor.respond(AbstractCompoundRequestCycleProcessor.java
>>>>:76)
>>>> wicket.RequestCycle.step(RequestCycle.java:1006)
>>>> wicket.RequestCycle.steps(RequestCycle.java:1040)
>>>>
 wicket.RequestCycle.request(RequestCycle.java:454)
>>>> wicket.protocol.http.WicketServlet.doGet(WicketServlet.java
>>>>:216)
>>>> javax.servlet.http.HttpServlet.service(HttpServlet.java:689)
>>>> javax.servlet.http.HttpServlet.service(HttpServlet.java:802)

>>>>Is there some config missing?

>>>>Thanks,

>>>>Jaime.

---

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