JSP页面内容如下，只是一个最简单的链接传递参数给Servlet
<%@ page contentType="text/html; charset=gb2312" language="java" import="java.util.*" errorPage="" %>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=gb2312" />
<title>Servlet接收表单示例</title>
</head>
<body>
<body>
<a href="/AcceptForm?name=Bill">调用Servlet,并传递参数</a>
</body>
</body>
</html>
Servlet的AcceptForm放在内容：ROOT\WEB-INF\classes\servlets下package servlets;
import java.io.*;
import javax.servlet.*;
import javax.servlet.http.*;
public class AcceptForm extends HttpServlet {
public void doGet(HttpServletRequest request, HttpServletResponse response)
throws IOException, ServletException {
response.setContentType("text/html");
response.setCharacterEncoding("gb2312");
PrintWriter out = response.getWriter();
String name = request.getParameter("name");
String province = request.getParameter("province"); out.println("<font size='2'>");
out.print("提交的表单内容为:<br>");
out.print("姓名:" + name + "<br>");
out.print("省份:" + province + "<br>");
out.print("</font>"); } public void doPost(HttpServletRequest request, HttpServletResponse response)
throws IOException, ServletException {
doGet(request, response);
}
}内容肯定是没错错误的，因为已经编译过了。
web.xml配置内容如下：
<servlet>
<servlet-name>AcceptForm</servlet-name>
<servlet-class>servlets.AcceptForm</servlet-class>
</servlet><servlet-mapping>
<servlet-name>AcceptForm</servlet-name>
<url-pattern>/AcceptForm</url-pattern>
</servlet-mapping>环境变量：
JAVA_HOME=C:\j2sdk1.4.2_04
Tomcat_HOME=C:\Tomcat5.0
CLASSPATH=.;%JAVA_HOME%\lib\tools.jar;%JAVA_HOME%\lib\dt.jar;%JDBC_LIB%\msbase.jar;%JDBC_LIB%\mssqlserver.jar;%JDBC_LIB%\msutil.jar;%Tomcat_HOME%\common\lib\servlet-api.jar;
Path=（前面的其他省略）;%JAVA_HOME%\bin;%Tomcat_HOME%\bin;
CATALINA_BASE=C:\Tomcat5.0
CATALINA_HOME=C:\Tomcat5.0
访问地址为：http://localhost/AcceptForm?name=Bill
报错内容如下：HTTP Status 404 - /AcceptForm--------------------------------------------------------------------------------type Status reportmessage /AcceptFormdescription The requested resource (/AcceptForm) is not available.
--------------------------------------------------------------------------------Apache Tomcat/5.0.28
请高手帮我一下，谢谢
<%@ page contentType="text/html; charset=gb2312" language="java" import="java.util.*" errorPage="" %>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=gb2312" />
<title>Servlet接收表单示例</title>
</head>
<body>
<body>
<a href="/AcceptForm?name=Bill">调用Servlet,并传递参数</a>
</body>
</body>
</html>
Servlet的AcceptForm放在内容：ROOT\WEB-INF\classes\servlets下package servlets;
import java.io.*;
import javax.servlet.*;
import javax.servlet.http.*;
public class AcceptForm extends HttpServlet {
public void doGet(HttpServletRequest request, HttpServletResponse response)
throws IOException, ServletException {
response.setContentType("text/html");
response.setCharacterEncoding("gb2312");
PrintWriter out = response.getWriter();
String name = request.getParameter("name");
String province = request.getParameter("province"); out.println("<font size='2'>");
out.print("提交的表单内容为:<br>");
out.print("姓名:" + name + "<br>");
out.print("省份:" + province + "<br>");
out.print("</font>"); } public void doPost(HttpServletRequest request, HttpServletResponse response)
throws IOException, ServletException {
doGet(request, response);
}
}内容肯定是没错错误的，因为已经编译过了。
web.xml配置内容如下：
<servlet>
<servlet-name>AcceptForm</servlet-name>
<servlet-class>servlets.AcceptForm</servlet-class>
</servlet><servlet-mapping>
<servlet-name>AcceptForm</servlet-name>
<url-pattern>/AcceptForm</url-pattern>
</servlet-mapping>环境变量：
JAVA_HOME=C:\j2sdk1.4.2_04
Tomcat_HOME=C:\Tomcat5.0
CLASSPATH=.;%JAVA_HOME%\lib\tools.jar;%JAVA_HOME%\lib\dt.jar;%JDBC_LIB%\msbase.jar;%JDBC_LIB%\mssqlserver.jar;%JDBC_LIB%\msutil.jar;%Tomcat_HOME%\common\lib\servlet-api.jar;
Path=（前面的其他省略）;%JAVA_HOME%\bin;%Tomcat_HOME%\bin;
CATALINA_BASE=C:\Tomcat5.0
CATALINA_HOME=C:\Tomcat5.0
访问地址为：http://localhost/AcceptForm?name=Bill
报错内容如下：HTTP Status 404 - /AcceptForm--------------------------------------------------------------------------------type Status reportmessage /AcceptFormdescription The requested resource (/AcceptForm) is not available.
--------------------------------------------------------------------------------Apache Tomcat/5.0.28
请高手帮我一下，谢谢
访问地址为：http://localhost/AcceptForm?name=Bill 你的工程名呢？http://localhost：8080（端口）/工程名/AcceptForm?name=Bill
改为
http://localhost：8080（端口）/工程名/AcceptForm?name=Bill