启动服务器时出现以下错误
localhost上的服务器Tomcat v7.0服务器无法启动。
当我查看控制台中的错误时,我认为这是问题"名为[Myclass]和[mypropackage.Myclass]的servlet都映射到url-pattern [/ myclass],这是不允许的& #34;
这里有什么问题以及如何解决?
Web.xml代码
<?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>myproject</display-name>
<welcome-file-list>
<welcome-file>index.html</welcome-file>
<welcome-file>index.htm</welcome-file>
<welcome-file>index.jsp</welcome-file>
<welcome-file>default.html</welcome-file>
<welcome-file>default.htm</welcome-file>
<welcome-file>default.jsp</welcome-file>
</welcome-file-list>
<servlet>
<servlet-name>Myclass</servlet-name>
<servlet-class>mypropackage.Myclass</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Myclass</servlet-name>
<url-pattern>/myclass</url-pattern>
</servlet-mapping>
</web-app>
Servlet -class代码:
package mypropackage;
import java.io.IOException;
import java.io.PrintWriter;
import javax.servlet.ServletConfig;
import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
/**
* Servlet implementation class Myclass
*/
@WebServlet("/myclass")
public class Myclass extends HttpServlet {
private static final long serialVersionUID = 1L;
/**
* @see HttpServlet#HttpServlet()
*/
public Myclass() {
super();
// TODO Auto-generated constructor stub
}
/**
* @see Servlet#init(ServletConfig)
*/
public void init(ServletConfig config) throws ServletException {
// TODO Auto-generated method stub
}
/**
* @see HttpServlet#service(HttpServletRequest request,
HttpServletResponse response)
*/
protected void service(HttpServletRequest request,
HttpServletResponse response) throws ServletException, IOException {
System.out.println("service method");
String uname = "user";
String pass = "abcd";
String un = request.getParameter("username");
String pw = request.getParameter("password");
String msg = " ";
if(un.equals(uname) && pw.equals(pass))
{
msg = "hello" + un + "login successful";
}
else
{
msg = "hello" + pw + "login failed";
}
response.setContentType("text/html");
PrintWriter out = response.getWriter();
out.println(msg);
}
}
这是登录控制台:
Caused by: java.lang.IllegalArgumentException: The servlets named [Myclass] and [mypropackage.Myclass] are both mapped to the url-pattern [/myclass] which is not permitted
at org.apache.catalina.deploy.WebXml.addServletMapping(WebXml.java:293)
at org.apache.catalina.startup.ContextConfig.processAnnotationWebServlet(ContextConfig.java:2428)
at org.apache.catalina.startup.ContextConfig.processAnnotationsStream(ContextConfig.java:2103)
at org.apache.catalina.startup.ContextConfig.processAnnotationsFile(ContextConfig.java:2064)
at org.apache.catalina.startup.ContextConfig.processAnnotationsFile(ContextConfig.java:2057)
at org.apache.catalina.startup.ContextConfig.webConfig(ContextConfig.java:1304)
at org.apache.catalina.startup.ContextConfig.configureStart(ContextConfig.java:889)
at org.apache.catalina.startup.ContextConfig.lifecycleEvent(ContextConfig.java:386)
at org.apache.catalina.util.LifecycleSupport.fireLifecycleEvent(LifecycleSupport.java:117)
at org.apache.catalina.util.LifecycleBase.fireLifecycleEvent(LifecycleBase.java:90)
at org.apache.catalina.core.StandardContext.startInternal(StandardContext.java:5412)
at org.apache.catalina.util.LifecycleBase.start(LifecycleBase.java:150)
... 6 more
答案 0 :(得分:2)
你确实两次定义了你的servlet:
将@WebServlet("/myclass")
注释用于课程mypropackage.MyClass
使用web.xml
定义servlet Myclass
。
servlet的两个实例都映射到同一个URL,这就是容器告诉你的。您应该只使用其中一种方法来定义servlet:使用注释或使用web.xml
。我更喜欢使用web.xml
,但这可能是因为我曾经在注释发明之前的几年定义了servelts。所以,欢迎您自己做出选择。
使用web.xml的IMHO更灵活。您无需确定在开发时映射servlet的URL是什么,并且可以多次部署相同的servlet并将其映射到多个URL。