的index.html
<!DOCTYPE html>
<html>
<head>
<meta charset="ISO-8859-1">
<title>sendRedirect example</title>
</head>
<body>
<form action="MySearcher">
<input type="text" name="name">
<input type="submit" value="Google Search">
</form>
</body>
</html>
MySearcher.java
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 MySearcher extends HttpServlet {
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
String name=request.getParameter("name");
response.sendRedirect("https://www.google.co.in/#q="+name);
}
}
web.xml中
<?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_2_5.xsd" id="WebApp_ID" version="2.5">
<display-name>GoogleSearcher</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>
<description></description>
<display-name>MySearcher</display-name>
<servlet-name>MySearcher</servlet-name>
<servlet-class>MySearcher</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>MySearcher</servlet-name>
<url-pattern>/MySearcher</url-pattern>
</servlet-mapping>
</web-app>
我尝试使用servlet在eclipse中创建一个程序,用户将进入他想要搜索的内容,然后他将被引导到google上的搜索结果.程序在eclipse上运行.但有一点问题.我没有被引导到搜索结果页面.它指示我到谷歌主页.有谁能告诉我这个问题的原因,?? 帮助将不胜感激.我只是一个初学者,试图学习servlet和我的第二个程序
您需要更改此行:
response.sendRedirect("https://www.google.co.in/#q="+name);
至
response.sendRedirect("https://www.google.co.in/search?q="+name);
这是@BalusC编译信息的链接
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