我是servlet的新手。我正在低于sevlets例外。
我的web.xml代码。
<display-name>LoginServlets</display-name>
<servlet>
<servlet-name>LoginServlet</servlet-name>
<servlet-class>LoginServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>LoginServlet</servlet-name>
<url-pattern>/login</url-pattern>
</servlet-mapping>
<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>
调用此servlet的Html表单。
<body>
<form action="login" method="get">
<table>
<tr>
<td>UserName</td>
<td><input type="text" name="username" /></td>
</tr>
<tr>
<td>Password</td>
<td><input type="password" name="password" /></td>
</tr>
<tr>
<td><input type="submit" value="SignIn" /></td>
</tr>
</table>
</form>
</body>
我得到了例外。
javax.servlet.ServletException: Error instantiating servlet class LoginServlet
org.apache.catalina.authenticator.AuthenticatorBase.invoke(AuthenticatorBase.java:501)
org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:103)
org.apache.catalina.valves.AccessLogValve.invoke(AccessLogValve.java:950)
org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:408)
org.apache.coyote.http11.AbstractHttp11Processor.process(AbstractHttp11Processor.java:1070)
org.apache.coyote.AbstractProtocol$AbstractConnectionHandler.process(AbstractProtocol.java:611)
org.apache.tomcat.util.net.JIoEndpoint$SocketProcessor.run(JIoEndpoint.java:314)
java.util.concurrent.ThreadPoolExecutor.runWorker(Unknown Source)
java.util.concurrent.ThreadPoolExecutor$Worker.run(Unknown Source)
org.apache.tomcat.util.threads.TaskThread$WrappingRunnable.run(TaskThread.java:61)
java.lang.Thread.run(Unknown Source)
根本原因。
java.lang.ClassNotFoundException: LoginServlet
org.apache.catalina.loader.WebappClassLoader.loadClass(WebappClassLoader.java:1720)
org.apache.catalina.loader.WebappClassLoader.loadClass(WebappClassLoader.java:1571)
org.apache.catalina.authenticator.AuthenticatorBase.invoke(AuthenticatorBase.java:501)
org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:103)
org.apache.catalina.valves.AccessLogValve.invoke(AccessLogValve.java:950)
org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:408)
org.apache.coyote.http11.AbstractHttp11Processor.process(AbstractHttp11Processor.java:1070)
org.apache.coyote.AbstractProtocol$AbstractConnectionHandler.process(AbstractProtocol.java:611)
org.apache.tomcat.util.net.JIoEndpoint$SocketProcessor.run(JIoEndpoint.java:314)
java.util.concurrent.ThreadPoolExecutor.runWorker(Unknown Source)
java.util.concurrent.ThreadPoolExecutor$Worker.run(Unknown Source)
org.apache.tomcat.util.threads.TaskThread$WrappingRunnable.run(TaskThread.java:61)
java.lang.Thread.run(Unknown Source)
去年类似的代码工作正常。但我也在使用git SVN我用相同的代码设置了一个创建新项目。但它不起作用。我也下载了一些样品但没有工作。
这是我的LoginServlet类。它位于JavaResources-&gt; src-mypackage
中package com.loginservlets;
@WebServlet("/LoginServlet")
public class LoginServlet extends HttpServlet {
private static final long serialVersionUID = 1L;
/**
* @see HttpServlet#HttpServlet()
*/
public LoginServlet() {
super();
// TODO Auto-generated constructor stub
System.out.println("LoginServlet");
}
/**
* @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse
* response)
*/
protected void doGet(HttpServletRequest request,
HttpServletResponse response) throws ServletException, IOException {
// TODO Auto-generated method stub
System.out.println("Get");
response.setContentType("text/html");
PrintWriter pw = response.getWriter();
pw.println("HelloWorld");
out.close();
}
/**
* @see HttpServlet#doPost(HttpServletRequest request, HttpServletResponse
* response)
*/
protected void doPost(HttpServletRequest request,
HttpServletResponse response) throws ServletException, IOException {
System.out.println("Post");
}
}
答案 0 :(得分:5)
您班级的名称是com.loginservlets.LoginServlet
。不是LoginServlet
。
请注意,在web.xml中声明servlet是多余的,并使用注释声明它。没有理由在web.xml中声明它,因为你可以对注释做同样的事情:
@WebServlet("/login")
答案 1 :(得分:0)
由于您已使用@WebServlet
注释了servlet类定义,因此不需要在web.xml
中声明相同的内容。将@WebServlet("/LoginServlet")
更改为@WebServlet(urlPatterns = {"/login"})