Question

tomcat下部署web项目从页面跳转至servlet的404错误，求指点！在线等啊

web.xml:
<servlet>
<servlet-name>login</servlet-name>
<servlet-class>com.test.servlet.LoginServlet</servlet-class>
</servlet>

<servlet-mapping>
<servlet-name>login</servlet-name>
<url-pattern>/loginServlet</url-pattern>
</servlet-mapping>
页面表单部分：
<form id="form1" name="form1" method="post" action="/loginServlet">
<table>
<tr>
<td>用户名：

<label>
<input type="text" name="username" />
</label>

</td>
</tr>
<tr>
<td>密 码：
<input type="password" name="password" />
</td>
</tr>
<tr>
<td><label>
<input type="submit" name="Submit" value="提交" />

<input type="reset" name="Submit2" value="重置" />
</label></td>
</tr>
</table>
</form>

servlet：
public class LoginServlet extends HttpServlet
{
private static final long serialVersionUID = 1L;

/**
* Default constructor.
*/
public LoginServlet()
{

}

/**
*
*/
protected void doGet(HttpServletRequest request,
HttpServletResponse response) throws ServletException, IOException
{
this.doPost(request, response);
}

/**
*
*/
protected void doPost(HttpServletRequest request,
HttpServletResponse response) throws ServletException, IOException
{
PrintWriter writer = response.getWriter();
// writer.print("<script>alert('登录成功!');window.location.href='跳到登陆成功后的页面'</script>");
writer.print("<script>alert('登录成功!');</script>");

String username = request.getParameter("username");
String password = request.getParameter("password");
System.out.println(username + "," + password);
request.getRequestDispatcher("/success.jsp").forward(request, response);
// response.sendRedirect("/WebTest/WebContent/jsp/success.jsp");

}

Answer 1

<form id="form1" name="form1" method="post" action="/loginServlet">\

<servlet-name>login</servlet-name>

看见这俩个了么？
你提交的是loginServlet，但是你的web.xml中没有这个servlet，
也就是你在配置这个xml文件的过程中出现了错误。

改成login试试！