如何使用Web.xml中的错误页面处理RESTWeb服务Java中的servlet异常?
我想在rest web服务中处理Servlet异常。我正在web.xml文件中使用error pages标记,但仍然无法捕获异常。我应该将ErrorHandler.java文件放在哪里。下面是我的web.xml文件如何使用Web.xml中的错误页面处理RESTWeb服务Java中的servlet异常?,java,web-services,spring-mvc,web.xml,servletexception,Java,Web Services,Spring Mvc,Web.xml,Servletexception,我想在rest web服务中处理Servlet异常。我正在web.xml文件中使用error pages标记,但仍然无法捕获异常。我应该将ErrorHandler.java文件放在哪里。下面是我的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 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>
Web应用程序创建的原型
调度员
org.springframework.web.servlet.DispatcherServlet
调度员
/
错误处理程序
错误处理程序
错误处理程序
/错误处理程序
404
/错误处理程序
java.lang.Throwable
/错误处理程序
log4j配置位置
WEB-INF/classes/log4j.properties
下面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);
}
}
//导入所需的java库
导入java.io.*;
导入javax.servlet.*;
导入javax.servlet.http.*;
导入java.util.*;
//扩展HttpServlet类
公共类ErrorHandler扩展了HttpServlet{
//方法来处理GET方法请求。
public void doGet(HttpServletRequest请求、HttpServletResponse响应)
抛出ServletException、IOException{
//分析servlet异常
可丢弃可丢弃=(可丢弃)
getAttribute(“javax.servlet.error.exception”);
整数状态代码=(整数)
getAttribute(“javax.servlet.error.status_code”);
字符串servletName=(字符串)
getAttribute(“javax.servlet.error.servlet_name”);
if(servletName==null){
servletName=“未知”;
}
字符串requestUri=(字符串)
getAttribute(“javax.servlet.error.request_uri”);
if(requestUri==null){
requestUri=“未知”;
}
//设置响应内容类型
response.setContentType(“text/html”);
PrintWriter out=response.getWriter();
String title=“错误/异常信息”;
字符串docType=
“\n”;
out.println(docType+
“\n”+
“”+标题+“\n”+
“\n”);
if(throwable==null&&statusCode==null){
out.println(“错误信息丢失”);
out.println(“请返回到。”);
}else if(状态代码!=null){
out.println(“状态代码:+statusCode”);
}否则{
out.println(“错误信息”);
out.println(“Servlet名称:“+servletName+””;
println(“异常类型:“+throwable.getClass().getName()+””;
println(“请求URI:+requestUri+”
);
println(“异常消息:+throwable.getMessage());
}
out.println(“”);
out.println(“”);
}
//方法来处理POST方法请求。
public void doPost(HttpServletRequest请求、HttpServletResponse响应)
抛出ServletException、IOException{
doGet(请求、响应);
}
}
我应该把上面的java文件放在哪里?
我必须手动将其编译为类文件并放入/webapps/ROOT/WEB-INF/classes目录。请告诉我正确的处理方法。我不知道REST服务与UI有什么关系。它们应该完全解耦。REST服务应该捕获任何异常并创建适当的响应以返回给用户。如果这恰好是在错误页面中呈现的,那就这样吧。您应该将Java文件放在
src/main/Java/{package\u nname}
如何在servlet异常中返回自定义错误xml响应?