Java TOMCAT状态404尝试了所有错误
我试图在tomcat上运行web.xml文件并在其中设置服务器/服务器映射。显然,错误日志似乎没有错误(如果我没有错的话),我试着用localhost:8080/Example03/servlet1、localhost:8080/Example03和 localhost:8080/servlet1当我读到一些文章时,在某些tomcat版本上发布了新的实现。。仅显示http错误404(状态报告-请求的资源不可用)。使用v6.0。。有人知道我能做些什么吗Java TOMCAT状态404尝试了所有错误,java,apache,http,tomcat,http-status-code-404,Java,Apache,Http,Tomcat,Http Status Code 404,我试图在tomcat上运行web.xml文件并在其中设置服务器/服务器映射。显然,错误日志似乎没有错误(如果我没有错的话),我试着用localhost:8080/Example03/servlet1、localhost:8080/Example03和 localhost:8080/servlet1当我读到一些文章时,在某些tomcat版本上发布了新的实现。。仅显示http错误404(状态报告-请求的资源不可用)。使用v6.0。。有人知道我能做些什么吗 Jan 08, 2014 5:59:48 P
Jan 08, 2014 5:59:48 PM org.apache.catalina.core.AprLifecycleListener init
INFO: The APR based Apache Tomcat Native library which allows optimal performance in production environments was not found on the java.library.path: C:\Program Files\Java\jre7\bin;C:\Windows\Sun\Java\bin;C:\Windows\system32;C:\Windows;C:\Windows\system32;C:\Windows;C:\Windows\System32\Wbem;C:\Windows\System32\WindowsPowerShell\v1.0\;C:\Program Files (x86)\Windows Live\Shared;.
Jan 08, 2014 5:59:48 PM org.apache.tomcat.util.digester.SetPropertiesRule begin
WARNING: [SetPropertiesRule]{Server/Service/Engine/Host/Context} Setting property 'source' to 'org.eclipse.jst.jee.server:Example03' did not find a matching property.
Jan 08, 2014 5:59:48 PM org.apache.coyote.http11.Http11Protocol init
INFO: Initializing Coyote HTTP/1.1 on http-8080
Jan 08, 2014 5:59:48 PM org.apache.catalina.startup.Catalina load
INFO: Initialization processed in 494 ms
Jan 08, 2014 5:59:48 PM org.apache.catalina.core.StandardService start
INFO: Starting service Catalina
Jan 08, 2014 5:59:48 PM org.apache.catalina.core.StandardEngine start
INFO: Starting Servlet Engine: Apache Tomcat/6.0.26
Jan 08, 2014 5:59:48 PM org.apache.coyote.http11.Http11Protocol start
INFO: Starting Coyote HTTP/1.1 on http-8080
Jan 08, 2014 5:59:48 PM org.apache.jk.common.ChannelSocket init
INFO: JK: ajp13 listening on /0.0.0.0:8009
Jan 08, 2014 5:59:48 PM org.apache.jk.server.JkMain start
INFO: Jk running ID=0 time=0/32 config=null
Jan 08, 2014 5:59:48 PM org.apache.catalina.startup.Catalina start
INFO: Server startup in 272 ms
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" 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">
<servlet>
<servlet-name>Test</servlet-name>
<servlet-class>org.test.webapp.ServletExample</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Test</servlet-name>
<url-pattern>/servlet1</url-pattern>
</servlet-mapping>
<display-name>Example03</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>
</web-app>
类别:
package org.test.webapp;
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 ServletExample extends HttpServlet {
public void doGet(HttpServletRequest req, HttpServletResponse resp)
throws ServletException,IOException{
String name = req.getParameter("userName");
String email = req.getParameter("email");
String IP = req.getRemoteAddr();
resp.getWriter().println("<html>");
resp.getWriter().println("<head>");
resp.getWriter().println("<title>this is the response</title>");
resp.getWriter().println("</head>");
resp.getWriter().println("<body>");
resp.getWriter().println("your name is: "+ name);
resp.getWriter().println("your email is: "+ email);
resp.getWriter().println("your ip is: "+ IP);
resp.getWriter().println("</body>");
resp.getWriter().println("</html>");
}
}
package org.test.webapp;
导入java.io.IOException;
导入javax.servlet.ServletException;
导入javax.servlet.http.HttpServlet;
导入javax.servlet.http.HttpServletRequest;
导入javax.servlet.http.HttpServletResponse;
公共类ServletExample扩展了HttpServlet{
公共无效数据集(HttpServletRequest请求、HttpServletResponse响应)
抛出ServletException、IOException{
字符串名称=req.getParameter(“用户名”);
字符串email=req.getParameter(“email”);
字符串IP=req.getRemoteAddr();
resp.getWriter()println(“”);
resp.getWriter()println(“”);
resp.getWriter().println(“这是响应”);
resp.getWriter()println(“”);
resp.getWriter()println(“”);
resp.getWriter().println(“您的名字是:”+name);
resp.getWriter().println(“您的电子邮件是:“+email”);
resp.getWriter().println(“您的ip是:“+ip”);
resp.getWriter()println(“”);
resp.getWriter()println(“”);
}
}
WEB-INF中的index.html:
<html>
<head>
<title> Test Form</title>
</head>
<body>
<form action="servlet1" method="get">
Name : <input type="text" name="name"><br>
Email : <input type="text" name="email"><br>
<input type = "submit" value="submit!">
</form>
</body>
</html>
测试表
名称:
电子邮件:
根据您的评论,我建议您从头开始使用tomcat。看看这个
只是简单介绍一下tomcat的工作原理
$Tomcat_Dir$/bin
并执行startup.sh
或startup.bat
(对于基于windows的系统)localhost:8080
应该为您提供tomcat的默认页面myApp.war
放在$Tomcat\u Dir$/webapps
文件夹中,然后重新启动Tomcat。您将看到在webapps文件夹中创建的文件夹与war同名http://localhost:8080/myApp
将打开应用程序的默认页面(如果已配置)。否则,根据web.xml,localhost:8080/myApp/servlet1
将启动servlet您正在部署任何应用程序?如果您正在部署任何应用程序,请发布该应用程序的
web.xml
。是的,刚刚添加了web.xml。它应该在屏幕上显示用户名和密码项。在WebContent中还有一个.html文件。您的Example03文件夹中是否有任何文件默认值。*或索引。*?没有。在文件夹。你的日志根本没有显示正在部署的应用程序。你是如何部署它的?呵呵,是的,没问题。最后一个问题。我需要下载core和deployer来运行web应用程序吗?你只需要core
就可以了
<html>
<head>
<title> Test Form</title>
</head>
<body>
<form action="servlet1" method="get">
Name : <input type="text" name="name"><br>
Email : <input type="text" name="email"><br>
<input type = "submit" value="submit!">
</form>
</body>
</html>