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电梯直达

1楼 [收藏(0)] [报告]

发表于 2005-03-28 17:13 |只看该作者 |倒序浏览

我的主要的问题是不知道怎么用表单提交给servlet来处理，我的问提如下：
1，首先我在建了一个Form1.html文件，结构如下：
E:\java\Tomcat 5.0\webapps\examples
         ——————————————— +WEB-INF
                     -----------------------------------+ classes
                                 -----------------------------------Form1.class
                     -------------------------------------web.xml
   ------------------------------------ Form.html
Form1.html文件内容如下：
<html>;
<head>;
<title>;Form.jsp</title>;
</head>;
<center>;
<body bgcolor="#FFFFFF">;
<form method="get" action="http://localhost:8080/examples/Form1" >;
<p>;
name:<input type="text" size="10" name="name">;
<p>;
email:<input type="text" size="10" name="email">;
<p>;
<input type="submit" name="submit" value="ok">;
</form>;
</body>;
</center>;
</html>;
2，Form1.java文件如下：
import java.io.*;
import java.util.*;
import javax.servlet.*;
import javax.servlet.http.*;
public class Form1 extends HttpServlet
{
public void doSet(HttpServletRequest rs,HttpServletResponse rp)
throws IOException,ServletException
{
rp.setContentType("text/html"

;
PrintWriter out = rp.getWriter();
out.println("<html>;"

;
out.println("<head>;"

;
out.println("<title>;"+"ha"+"</title>;"

;
out.println("</head>;"

;
out.println("<body>;"

;

Enumeration e = rs.getParameterNames();
while(e.hasMoreElements())
{
String key = (String)e.nextElement();
String value = rs.getParameter(key);
out.println("<p>;"

;
out.println(key+"="+value);
}
out.println("</body>;"

;
out.println("<html>;"

;
out.close();
}
}

3，web.xml文件如下：
<?xml version="1.0" encoding="ISO-8859-1"?>;

<!DOCTYPE web-app
PUBLIC "-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN"
"http://java.sun.com/dtd/web-app_2_3.dtd">;

<web-app>;
<display-name>;Welcome to Tomcat</display-name>;
<description>;
Welcome to Tomcat
</description>;
<servlet>;
<servlet-name>;Hello</servlet-name>;
<servlet-class>;Hello</servlet-class>;
</servlet>;

<servlet>;
<servlet-name>;Form1</servlet-name>;
<servlet-class>;Form1</servlet-class>;
</servlet>;

<servlet-mapping>;
<servlet-name>;Hello</servlet-name>;
<url-pattern>;/Hello</url-pattern>;
</servlet-mapping>;

<servlet-mapping>;
<servlet-name>;Form1</servlet-name>;
<url-pattern>;/Form1</url-pattern>;
</servlet-mapping>;

</web-app>;
4，我写的Hello的servlet可以运行，
5，现在的问题是：
下面是tomcat 的提示：
HTTP Status 405 - HTTP method GET is not supported by this URL

--------------------------------------------------------------------------------

type Status report

message HTTP method GET is not supported by this URL

description The specified HTTP method is not allowed for the requested resource (HTTP method GET is not supported by this URL).

--------------------------------------------------------------------------------

Apache Tomcat/5.0.28
谢谢了[/img]

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siqiabc

稍有积蓄

论坛徽章:: 0

2楼 [报告]

发表于 2005-03-28 17:30 |只看该作者

关于servlet的问题？？？？？

在<form action="servlet名字">;
然后可以在servlet处理form 表单提交过来的数据。
hen jian dan de

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yangguangwen

白手起家

论坛徽章:: 0

3楼 [报告]

发表于 2005-03-28 18:10 |只看该作者

关于servlet的问题？？？？？

我试过了，你说的方法不行！

servlet的名字"Form1"不行！原来的那个已不行！
我的问题是：
HTTP Status 405 - HTTP method GET is not supported by this URL

--------------------------------------------------------------------------------

type Status report

message HTTP method GET is not supported by this URL

description The specified HTTP method is not allowed for the requested resource (HTTP method GET is not supported by this URL).

--------------------------------------------------------------------------------

Apache Tomcat/5.0.28