j2ee中servlet的一个新手问题，页面跳转提示The requested resource is not available.

Question

<form action="servlet_in" method="post">
        <div class="hom-add-box">
            <label>用户名：</label> <input type="text" id="hom-in-username">
        </div>
        <div class="hom-add-box">
            <label>密码：</label> <input type="password" id="hom-in-password">
        </div>
        <div class="hom-add-box">
            <label>登陆类型：</label>
            <select>
                <option>用户</option>
                <option>管理员</option>
            </select>
        </div>
        <div class="hom-add-box">
            <input type="submit" value="登陆" id="submit" />
        </div>
    </form>

servlet_in中dopost是这样写的：

    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        // TODO Auto-generated method stub
        request.setCharacterEncoding("GB18030");
        String username=request.getParameter("hom-in-username");
        String password=request.getParameter("hom-in-password");
        request.setAttribute("name", username);
        request.setAttribute("password", password);
        request.getRequestDispatcher("hom.jsp").forward(request, response);
    }

xml配置：

<?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" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" id="WebApp_ID" version="3.0">
  <display-name>j2ee-demo</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>
    <servlet-name>servlet_in</servlet-name>
    <servlet-class>page.servlet_in</servlet-class>
  </servlet>
  <servlet-mapping>
    <servlet-name>servlet_in</servlet-name>
    <url-pattern>/servlet_in</url-pattern>
  </servlet-mapping>
</web-app>

Answer 1

form的action应该是page/servlet_in吧

Answer 2

书上说，<url–pattern>的格式如果是“/computerBill”那么action就是"computerBill"