小编War*_*007的帖子

我正在学习Java servlet,我为"GET"和"POST"编写了两个单独的servlet.当向服务器发送"GET"请求时,servlet访问数据库并检索所有内容并将结果转换为Google Charts可识别的格式.当向服务器发送"POST"请求时,servlet获取参数并将它们添加到Java对象,然后DAO将数据添加到数据库.但是,当我在输入后点击"添加"按钮时,Web应用程序根本找不到servlet.它只是"跳过"ajax函数并继续.所以这是插入的servlet:

@WebServlet("/InsertServlet")
public class InsertServlet extends HttpServlet 
{
    private static final long serialVersionUID = 1L;
    private EmployeeDao dao;

    public InsertServlet() throws SQLException 
    {
        super();
        dao = new EmployeeDao();
    }

    public void doPost(HttpServletRequest request, HttpServletResponse response) 
            throws ServletException, IOException 
    {
        System.out.println("doPost");
        Employee e = new Employee();
        e.setName(request.getParameter("name"));
        e.setSSN(request.getParameter("ssn"));
        e.setDob(request.getParameter("birth"));
        e.setIncome(request.getParameter("xxxx"));

        dao.addEmployee(e);

        response.setContentType("text/html;charset=utf-8");
        PrintWriter out = response.getWriter();
        out.println("<h2>Data Entry Added</h2><br>");
        out.println("<h2>"+request.getParameter("name")+"</h2>");
        out.println("<h2>"+request.getParameter("ssn")+"</h2>");
        out.println("<h2>"+request.getParameter("birth")+"</h2>");
        out.println("<h2>"+request.getParameter("xxxx")+"</h2>");
        out.flush();
        out.close();


    }
}

这是index.html:

<form id="inputForm">
<table style="width:80%;border:3px;">
    <tr>
        <td align="center"><input type="text" name="name" id="name" placeholder="First …