Question

我想在我的其他Web服务中处理Servlet异常。我在web.xml文件中使用了error-pages标记，但仍然无法捕获异常。我应该放置我的ErrorHandler.java文件.Below是我的web.xml文件

    <?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" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" id="WebApp_ID" version="3.0"> 


    <display-name>Archetype Created Web Application</display-name>

    <servlet>
        <servlet-name>dispatcher</servlet-name>
        <servlet-class>
            org.springframework.web.servlet.DispatcherServlet
        </servlet-class>
    </servlet>
    <servlet-mapping>
        <servlet-name>dispatcher</servlet-name>
        <url-pattern>/</url-pattern>
    </servlet-mapping>

<servlet>
   <servlet-name>ErrorHandler</servlet-name>
   <servlet-class>ErrorHandler</servlet-class>
</servlet>

<!-- servlet mappings -->
<servlet-mapping>
   <servlet-name>ErrorHandler</servlet-name>
   <url-pattern>/ErrorHandler</url-pattern>
</servlet-mapping>

<error-page>
   <error-code>404</error-code>
   <location>/ErrorHandler</location>
</error-page>

<error-page>
   <exception-type>java.lang.Throwable</exception-type >
   <location>/ErrorHandler</location>
</error-page>
    <context-param>
        <param-name>log4j-config-location</param-name>
        <param-value>WEB-INF/classes/log4j.properties</param-value>
    </context-param>
</web-app>

这是我的示例异常处理程序代码 -

// Import required java libraries
import java.io.*;
import javax.servlet.*;
import javax.servlet.http.*;
import java.util.*;

// Extend HttpServlet class
public class ErrorHandler extends HttpServlet {

   // Method to handle GET method request.
   public void doGet(HttpServletRequest request, HttpServletResponse response)
      throws ServletException, IOException {

      // Analyze the servlet exception       
      Throwable throwable = (Throwable)
      request.getAttribute("javax.servlet.error.exception");
      Integer statusCode = (Integer)
      request.getAttribute("javax.servlet.error.status_code");
      String servletName = (String)
      request.getAttribute("javax.servlet.error.servlet_name");

      if (servletName == null) {
         servletName = "Unknown";
      }
      String requestUri = (String)
      request.getAttribute("javax.servlet.error.request_uri");

      if (requestUri == null) {
         requestUri = "Unknown";
      }

      // Set response content type
      response.setContentType("text/html");

      PrintWriter out = response.getWriter();
      String title = "Error/Exception Information";
      String docType =
         "<!doctype html public \"-//w3c//dtd html 4.0 " +
         "transitional//en\">\n";

      out.println(docType +
         "<html>\n" +
         "<head><title>" + title + "</title></head>\n" +
         "<body bgcolor = \"#f0f0f0\">\n");

      if (throwable == null && statusCode == null) {
         out.println("<h2>Error information is missing</h2>");
         out.println("Please return to the <a href=\"" + 
            response.encodeURL("http://localhost:8080/") + 
            "\">Home Page</a>.");
      } else if (statusCode != null) {
         out.println("The status code : " + statusCode);
      } else {
         out.println("<h2>Error information</h2>");
         out.println("Servlet Name : " + servletName + "</br></br>");
         out.println("Exception Type : " + throwable.getClass( ).getName( ) + "</br></br>");
         out.println("The request URI: " + requestUri + "<br><br>");
         out.println("The exception message: " + throwable.getMessage( ));
      }
      out.println("</body>");
      out.println("</html>");
   }

   // Method to handle POST method request.
   public void doPost(HttpServletRequest request, HttpServletResponse response)
      throws ServletException, IOException {

      doGet(request, response);
   }
}

我应该在哪里放置上面的java文件？我手动将其编译为类文件并放在/ webapps / ROOT / WEB-INF / classes目录中。请告诉我处理它的正确方法。

如何使用web.xml中的错误页面处理Rest Web Services Java中的servlet异常？

0 个答案: