Я работаю с Spring MVC 3.2.6.RELEASE и WebLogic 10.3.6 и Eclipse, JunoSpring MVC WebLogic 10.3.6 Ошибка 403-Запретный
Когда я запустить приложение, я хочу пойти контроллер и получить JSP ,
Это мой web.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:jsp="http://java.sun.com/xml/ns/javaee/jsp" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" version="2.5">
<display-name>eusurveyadmin</display-name>
<servlet>
<servlet-name>eusurveyadmin</servlet-name>
<servlet-class>
org.springframework.web.servlet.DispatcherServlet
</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>
/WEB-INF/spring/eusurveyadmin-servlet.xml
</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>eusurveyadmin</servlet-name>
<url-pattern>/eusurveyadmin/*</url-pattern>
</servlet-mapping>
<jsp-config>
<taglib>
<taglib-uri>http://www.owasp.org/index.php/Category:OWASP_Enterprise_Security_API</taglib-uri>
<taglib-location>/WEB-INF/Content/esapi.tld</taglib-location>
</taglib>
</jsp-config>
<context-param>
<param-name>log4jConfigLocation</param-name>
<param-value>file:ecalcpAdminlog4j.xml</param-value>
</context-param>
<listener>
<listener-class>
eusurvey.listener.Log4jConfigListener</listener-class>
</listener>
<context-param>
<param-name>propertiesConfigLocation</param-name>
<param-value>file://C://OEPEJUNO//user_projects//domains//test38//configuracion.properties</param-value>
</context-param>
<listener>
<listener-class>
eusurvey.listener.CustomContextLoaderListener
</listener-class>
</listener>
<error-page>
<error-code>500</error-code>
<location>/500</location>
</error-page>
<error-page>
<error-code>404</error-code>
<location>/404</location>
</error-page>
</web-app>
Контроллер WelcomeController.java
package eusurvey.controller;
import java.io.PrintWriter;
import java.io.StringWriter;
import java.util.List;
import java.util.Locale;
import javax.annotation.Resource;
import javax.servlet.http.HttpServletRequest;
import javax.validation.Valid;
import org.apache.log4j.Logger;
import org.springframework.stereotype.Controller;
import org.springframework.ui.Model;
import org.springframework.ui.ModelMap;
import org.springframework.validation.BindingResult;
import org.springframework.web.bind.annotation.ExceptionHandler;
import org.springframework.web.bind.annotation.ModelAttribute;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;
import org.springframework.web.servlet.ModelAndView;
import org.springframework.web.servlet.mvc.support.RedirectAttributes;
import org.springframework.web.servlet.view.RedirectView;
import eusurvey.modelA.daos.Preferencia;
import eusurvey.services.PreferencesService;
@Controller
@RequestMapping("/welcome")
public class WelcomeController extends ExceptionsController {
private static final Logger logger = Logger
.getLogger(WelcomeController.class);
@Resource(name = "preferencesService")
private PreferencesService preferencesService;
private int a = 0;
private Preferencia results = null;
@ModelAttribute("Preferencia")
public Preferencia fechaUltimaEncuesta() {
results = preferencesService.consultaPreferencia();
return results;
}
@RequestMapping(value = "/*")
public String welcome(HttpServletRequest request, ModelMap model) {
logger.info("WelcomeController welcome");
results = fechaUltimaEncuesta();
model.addAttribute("fechaUltimaEncuesta", results.getValor());
request.getSession().setAttribute("fechaUltimaEncuesta",
results.getValor());
String fechaUltimaEncuesta = (String) request.getSession()
.getAttribute("fechaUltimaEncuesta");
//return "welcome1";
return "/menu/pantallaInicio";
}
@ExceptionHandler(Exception.class)
public ModelAndView handleException(Exception e, Locale locale, HttpServletRequest request) {
logger.error("WelcomeController handleException "+e.getLocalizedMessage()+" exception "+ e);
StringWriter sw = new StringWriter();
PrintWriter pw = new PrintWriter(sw);
e.printStackTrace(pw);
String mensajeException = sw.toString();
logger.error("WelcomeController handleException exception "+e.getClass().getSimpleName()+" mensaje "+mensajeException);
ModelAndView mav = new ModelAndView();
mav.addObject("exception", e);
mav.addObject("url", request.getRequestURL());
mav.setViewName("errores/errorGeneral");
return mav;
}
}
Когда я бегу applicacion я получаю ошибку
Это мой weblogic.xml
<?xml version="1.0" encoding="UTF-8"?>
<wls:weblogic-web-app
xmlns:wls="http://xmlns.oracle.com/weblogic/weblogic-web-app"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd http://xmlns.oracle.com/weblogic/weblogic-web-app http://xmlns.oracle.com/weblogic/weblogic-web-app/1.3/weblogic-web-app.xsd">
<wls:weblogic-version>10.3.6</wls:weblogic-version>
<wls:context-root>EcalpAdmin</wls:context-root>
<wls:container-descriptor>
<wls:prefer-web-inf-classes>true</wls:prefer-web-inf-classes>
</wls:container-descriptor>
</wls:weblogic-web-app>
Как мне написать свой web.xml для перехода к контроллеру?
localhost: 7001/EcalpAdmin не будет работать. EcalpAdmin не является вашим корневым контекстом – ScanQR