我正在尝试使用jsp，servlet，Tomcat-6.0.43和Eclipse构建Twitter搜索，并出现HTTP Status 404错误。任何人都可以检查我在哪里错了。我的代码：
first.jsp：
<%@ page language="java" contentType="text/html; charset=ISO-8859-1"
pageEncoding="ISO-8859-1"%>
<!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>
<form action="ServletValues.java" method="get">
Enter Twitter Search Details : <input type="text" name="first"><br>
<input type="submit">
</form>
</body>
</html>
TwitterServlet.java：
import java.io.IOException;
import java.util.List;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import twitter4j.Status;
import twitter4j.Twitter;
import twitter4j.TwitterException;
import twitter4j.TwitterFactory;
import twitter4j.auth.AccessToken;
public class TwitterServlet extends HttpServlet {
private static final long serialVersionUID = 1L;
public TwitterServlet() {
}
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
String CONSUMER_KEY = "[data]";
String CONSUMER_KEY_SECRET = "[data]";
String AccessToken = "[data]";
String AccessTokenSecret = "[data]";
response.setContentType("text/html");
String input1 = request.getParameter("first");
try{
Twitter twitter = new TwitterFactory().getInstance();
twitter.setOAuthConsumer(CONSUMER_KEY, CONSUMER_KEY_SECRET);
AccessToken oathAccessToken = new AccessToken(AccessToken, AccessTokenSecret);
twitter.setOAuthAccessToken(oathAccessToken);
List<Status> status = twitter.getUserTimeline(input1);
for (Status status2 : status)
{
System.out.println("---Tweet---"+status2.getText());
}}catch (TwitterException te){
System.out.println("Error occured "+te);
}
super.doPost(request, response);
}
}
web.xml：
<?xml version="1.0" encoding="UTF-8"?>
<web-app id="WebApp_ID" version="2.4" xmlns="http://java.sun.com/xml/ns/j2ee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
<display-name>
Twitter12</display-name>
<servlet>
<description>
</description>
<display-name>
TwitterServlet</display-name>
<servlet-name>TwitterServlet</servlet-name>
<servlet-class>
TwitterServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>TwitterServlet</servlet-name>
<url-pattern>/TwitterServlet</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>
</web-app>
错误：HTTP状态404-/Twitter12/ServletValues.java
类型状态报告
消息/Twitter12/ServletValues.java
描述所请求的资源不可用。
在您的jsp文件表单操作中，替换
<form action="ServletValues.java" method="get">
和
<form action="TwitterServlet" method="get">
因为在您web.xml的Servlet映射模式中TwitterServlet，您打算调用Servlet类。
同样在您的servlet中，您只是打印到System.out，而您应该使用
response.getOutputStream().write("---Tweet---"+status2.getText());
这样它就会显示在响应中
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