Question

1。 servlet文件

public class StudentController extends HttpServlet {
    private static final long serialVersionUID = 1L;

    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    System.out.println("inside servlet");

     dao dao=new daoimpl();
    String action=(request.getParameter("action"));
    if(action.equalsIgnoreCase("login")){
        System.out.println("inside login");
        Student student=new Student();
        System.out.println(request.getParameter("username"));


        student.setUsername(request.getParameter("username"));
        student.setPassword(request.getParameter("password"));
        ArrayList<Student>al=dao.getAllStudent(student);
        Gson gson=new Gson();
        //PrintWriter pw=response.getWriter();
            String json = gson.toJson(al);
            response.setContentType("application/json");
            response.setCharacterEncoding("UTF-8");
            response.getWriter().write(json);
    }

＆＃13;

index.jsp
<html>
<head>
   <script src="js/jquery.min.js" type="text/javascript" ></script>
  <script type="text/javascript">
$(document).ready(function() {
	$('#login').click(function(event) {
     var uname = $('#username').val();
        var pass = $('#pass').val();
        alert(uname+"-----"+pass);
        
    	alert("on click");
        $.post("Controller?action=login",{
                username : uname,
                password : pass
             },function(data) {
            	 alert("sucess");    
            	var json=$.parseJSON(data);
                  alert(json);
});
        });
});
  </script>
    
</head>
<body>
<form action="login" method="post">
User Name:<input type="text" name="uname" id="username"><br>
Password:<input type="text" name="password" id="pass"><br>
<input type="submit" value="Login" id="login">
<a href="registration.jsp">Register</a>
<h3 align="center">student Details</h3>
<div id="tablediv">
<table cellspacing="0" id="studenttable">
</table>
</div></form></body>
</html>

＆＃13;

甚至在代码中没有成功警报，请建议我提前感谢jquery的新功能..

Answer 1

试试这个

<script type="text/javascript">
$(document).ready(
  function() {
      $('form').submit(function(event) {
        event.preventDefault();

         var uname = $('#username').val();
            var pass = $('#pass').val();
            alert(uname+"-----"+pass);            
            alert("on submit");


            $.ajax(
              {
                url:"Controller?action=login",
                data: {username : uname,password : pass}
                type:'post',
                cache:false,
                timeout:8000
              }
            ).done(function(data){
                alert("sucess");    
                var json=$.parseJSON(data);
                alert(json);

            }).fail(function(jqxhr, textStatus, error){
              alert(textStatus);
              alert(error);
            });
      });
    }
  );
</script>

不调用使用JQuery成功函数的Ajax

1 个答案: