i_r*_*aqz 9 java eclipse tomcat servlets java-ee
我知道这是一个非常常见的问题,因为我在包括SO在内的多个论坛中发现了许多与此相关的问题.但我还没有找到解决方案我的web.xml(位于WEB-INF)
<?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: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" id="WebApp_ID" version="2.5">
<display-name>SMSProjectNew</display-name>
<welcome-file-list>
<welcome-file>index.html</welcome-file>
</welcome-file-list>
<servlet>
<description></description>
<display-name>ReceiveMessagesServlet</display-name>
<servlet-name>ReceiveMessagesServlet</servlet-name>
<servlet-class>com.sendreceive.ReceiveMessagesServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>ReceiveMessagesServlet</servlet-name>
<url-pattern>/ReceiveMessagesServlet</url-pattern>
</servlet-mapping>
</web-app>
html页面index.html,位于WebContent文件夹中
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
<title>Insert title here</title>
</head>
<body>
The application started successfully version 1:27
<form action="/ReceiveMessagesServlet" method="post">
<input type="text" name="number"/>
<input type="text" name="message"/>
<input type="submit" name="submit"/>
</form>
</body>
</html>
最后是servlet,ReceiveMessagesServlet,位于src\com.sendreceive包com.sendreceive中;
import java.io.IOException;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
public class ReceiveMessagesServlet extends HttpServlet {
private static final long serialVersionUID = 1L;
public ReceiveMessagesServlet() {
super();
// TODO Auto-generated constructor stub
}
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
processRequest(request,response);
}
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
processRequest(request,response);
}
protected void processRequest(HttpServletRequest request,
HttpServletResponse response) {
String responseMessage = request.getParameter("message");
String responseNumber = request.getParameter("number");
System.out.println(responseMessage+responseNumber);
}
}
我在eclipse中安装了tomcat插件.当我右键单击项目然后单击在服务器上运行项目时.tomcat服务器在eclipse中启动并显示index.html页面.但是当我在字段中输入一些值并单击submit..it给出了404错误..我从过去的2小时起一直在苦苦挣扎...请帮助..另外..我正在使用这个教程 http://www.ibm.com/developerworks/opensource/library/os-eclipse-tomcat/index.html
小智 11
由于action ="/ ReceiveMessagesServlet",您收到404错误,请删除斜杠.尝试使用action ="ReceiveMessagesServlet".
当您向URL模式添加斜杠时,容器将查找部署名称为"ReceiveMessagesServlet"的Web应用程序.由于这不存在,您将收到404错误.
当您将应用程序部署到servlet容器中时,您的URL可能会以上下文路径为前缀,该路径标识您的应用程序以及该容器中的其他应用程序(即/ReceiveMessagesServlet变为/MyApp/ReceiveMessagesServlet).
因此,您应该考虑这种可能性并相应地修改您的URL,例如,使用JSTL <c:url>:
<form action="<c:url = value = '/ReceiveMessagesServlet' />" method="post">
或者,没有JSTL:
<form action="${pageContext.request.contextPath}/ReceiveMessagesServlet" method="post">
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