servlet是要在WEB.XML里配置的, action里面就写你在XML里配置的路径
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<form action="registerServlet" method="post">
<input type="submit" value="注册"/>
</form>
RegisterServlet中
package com.j2ee.servlet;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 RegisterServlet extends HttpServlet {
public RegisterServlet(){
super();
}
@Override
public void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
// TODO Auto-generated method stub
System.out.println("RegisterServlet....");
} @Override
public void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
// TODO Auto-generated method stub
this.doGet(request, response);
} @Override
public void destroy() {
// TODO Auto-generated method stub
super.destroy();
} @Override
public void init() throws ServletException {
// TODO Auto-generated method stub
super.init();
}
}
在你的web.xml中配置文件如下: <servlet>
<servlet-name>RegisterServlet</servlet-name>
<servlet-class>com.j2ee.servlet.RegisterServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>RegisterServlet</servlet-name>
<url-pattern>/registerServlet</url-pattern>
</servlet-mapping>
看下web.xml中的配置是否正确
<description>This is the description of my J2EE component</description>
<display-name>This is the display name of my J2EE component</display-name>
<servlet-name>Register</servlet-name>
<servlet-class>com.j2ee.servlet.Register</servlet-class>
</servlet> <servlet-mapping>
<servlet-name>Register</servlet-name>
<url-pattern>/servlet/Register</url-pattern>
</servlet-mapping>
[/code]
action=“servlet/TestServlet”
这样你试试
<servlet>
<description>This is the description of my J2EE component </description>
<display-name>This is the display name of my J2EE component </display-name>
<servlet-name>Register </servlet-name>
<servlet-class>com.j2ee.servlet.Register </servlet-class>
</servlet> <servlet-mapping>
<servlet-name>Register </servlet-name>
<url-pattern>/servlet/Register </url-pattern>
</servlet-mapping> action=“servlet/TestServlet”
你得确定你已经发布了你的程序
<form method="GET" action="servlet/Register">
<input type="submit" value="tijiao"/>
</form>写错了,应该是action="servlet/Register "
这样就可以了
<input type="submit" value="tijiao"/>
</form>servlet在工程的src文件夹下,没有包,是个裸体类,名称是TestServlet内容是默认的:
配置文件是: <servlet>
<description>This is the description of my J2EE component</description>
<display-name>This is the display name of my J2EE component</display-name>
<servlet-name>TestServlet</servlet-name>
<servlet-class>TestServlet</servlet-class>
</servlet> <servlet-mapping>
<servlet-name>TestServlet</servlet-name>
<url-pattern>/servlet/TestServlet</url-pattern>
</servlet-mapping>
register.jsp就是不能访问.
我只能重新建立了一个register.jsp文件,这下子又可以了,看来不是代码问题..15楼的方法没有错,我了解了
谢谢大家的帮忙!
首先,你再新建文件类的时候就选择建立一个Servlet,然后Eclipse会自动的在web.xml中对servlet的路径进行配置,
然后在JSp页面中向指定的servlet跳转的时候可以这样
<form action="servlet/ServletName" method="post">
如果系统还是找不到的话,就这样
<form action="../servlet/ServletName" method="post">
这样肯定OK
祝你成功·~!~!