Question

我试图使用db连接登录表单。当一个类型错误的用户名ajax应该回复“错误的用户名”。我的servlet是用hibernate连接的，所以我可以用HPQL检查db的用户名。 Withouth Hibernate代码，我的servlet将字符串发回给我，没有任何问题。但是当我尝试检查用户名是否存在时，使用hibernate代码ajax无法回复任何响应。

我的index.html

<!DOCTYPE html>
<!--
To change this license header, choose License Headers in Project Properties.
To change this template file, choose Tools | Templates
and open the template in the editor.
-->
<html>
    <head>
        <title>pokusaj 2</title>
        <meta charset="UTF-8">
        <meta name="viewport" content="width=device-width, initial-scale=1.0">
         <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
         <script>
             function check(){
                 var username=document.getElementById("username").value;
                 var password=document.getElementById("password").value;

                 $.post("servletprovera", {"username":username, "password":password},
                 function(data){
                     document.getElementById("alert").innerHTML=data.result;
                 });//servletprovera is URL of my servlet
             }
         </script>
    </head>
    <body>
        <form >
            <input type="text" name="username" id="username"><br>
            <input type="password" name="password" id="password"><br>
            <b id="alert"></b><br>
            <input type="button" value="Login" onclick="check();">
        </form>
    </body>
</html>

我的servlet

@WebServlet(name = "ServletProvera", urlPatterns = {"/servletprovera"})
public class ServletProvera extends HttpServlet {

    /**
     * Processes requests for both HTTP <code>GET</code> and <code>POST</code>
     * methods.
     *
     * @param request servlet request
     * @param response servlet response
     * @throws ServletException if a servlet-specific error occurs
     * @throws IOException if an I/O error occurs
     */
    protected void processRequest(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {
        response.setContentType("application/json");
        String user=request.getParameter("username");

        Configuration myConf=new Configuration();
        myConf.configure("hibernate.cfg.xml");
        StandardServiceRegistry service=new StandardServiceRegistryBuilder().
                applySettings(myConf.getProperties()).build();
        SessionFactory myFactory=myConf.buildSessionFactory(service);
        Session conn=myFactory.openSession();
        Transaction t=conn.beginTransaction();
        List<Korisniciajax> list;//Korisniciajax-Hibernate_entity class
        list=conn.createQuery("k FROM Korisniciajax k WHERE k.korUser=\""+user+"\"").list();

        t.commit();
        conn.close();

        String result;

        if(list==null){
            result="wrong username";
        }
        else {
            result="success";
        }

         try (PrintWriter out = response.getWriter()) {
            /* TODO output your page here. You may use following sample code. */
            out.print("{\"result\":\""+result+"\"}");
        }



    }

    // <editor-fold defaultstate="collapsed" desc="HttpServlet methods. Click on the + sign on the left to edit the code.">
    /**
     * Handles the HTTP <code>GET</code> method.
     *
     * @param request servlet request
     * @param response servlet response
     * @throws ServletException if a servlet-specific error occurs
     * @throws IOException if an I/O error occurs
     */
    @Override
    protected void doGet(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {
        processRequest(request, response);
    }

    /**
     * Handles the HTTP <code>POST</code> method.
     *
     * @param request servlet request
     * @param response servlet response
     * @throws ServletException if a servlet-specific error occurs
     * @throws IOException if an I/O error occurs
     */
    @Override
    protected void doPost(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {
        processRequest(request, response);
    }

    /**
     * Returns a short description of the servlet.
     *
     * @return a String containing servlet description
     */
    @Override
    public String getServletInfo() {
        return "Short description";
    }// </editor-fold>

}

Answer 1

我解决了这个问题。问题出现在HPQL中。

我写了

list=conn.createQuery("k FROM Korisniciajax k WHERE k.korUser=\""+user+"\"").list();

我应该使用＆＃39;＆＃39;而不是＆＃34;＆＃34;

list=conn.createQuery(" SELECT k FROM Korisniciajax k WHERE korUser='"+user+"'").list();

使用ajax和hibernate登录表单

1 个答案: