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问题:

I have used ajax with jquery and my ajax url is calling another jsp page which submits form data to database the problem is whenever I'm running the insert query, its replying "Null values can't be inserted to database" kinda exception, but whenever I'm reloading same page and clicking submit button, its getting submitted to database. I think the problem is with something like postback method, I'm new to ajax so please need your help...

<html>
<head>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4/jquery.min.js"></script>
<script type="text/javascript">
    $(document).ready(function() {
        $('#save').click(function ()
        {
            $.ajax({
                type: "post",
                url: "test.jsp", //this is my servlet
                data: "input=" +$('#id').val()+"&output="+$('#op').val(),
                success: function(msg){      
                        $('#output').append(msg);
                }
            });
        });

    });
</script>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>JSP Page</title>
</head>
<body>
    input:<input id="id" type="text" name="" value="" /><br></br>
    output:<input id="op" type="text" name="" value="" /><br></br>
    <input type="button" value="save" name="save" id="save"/>
   <div id="output"></div>
</body>

JSP:

<%@page import ="com.connectionUtil.ConnectionUtil"%> 
<!DOCTYPE html> 
<script src="js/jquery.min.js" type="text/javascript"></script> 
<% String id = request.getParameter("input");
    try {
        Connection con = ConnectionUtil.getConnection();
        String sql = "insert into test(ID,...) values ('" + id + "',...)";
        PreparedStatement ps = con.prepareStatement(sql);
        ps.executeUpdate();
    } catch (SQLException sqe) {
        out.println(sqe);
    }%> 
<body> <h4><%=id%> is stored </h4> </body> </html>

回答1:

Your ajax code is wrong. It should be like this:

$.ajax({
  type: "post",
  url: "test.jsp", //this is my servlet
  data: {input : $('#id').val(), output: $('#op').val()},
  success: function(msg){      
      $('#output').append(msg);
  }
});

And then use the request.getParameter("input") to get the parameter value

UPDATE : Please note that i missed a comma (,) at the end of the line data:....

UPDATE 2 : This is a working demo of your code without the connections to db...

index.jsp:

<%@page contentType="text/html" pageEncoding="UTF-8"%>
<html>
<head>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4/jquery.min.js"></script>
<script type="text/javascript">
    $(document).ready(function() {
        $('#save').click(function ()
        {
            $.ajax({
                type: "post",
                url: "test.jsp", //this is my servlet
                data: {
                    input: $('#id').val(), 
                    output: $('#op').val()
                },
                success: function(msg){      
                        $('#output').append(msg);
                }
            });
        });

    });
</script>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>JSP Page</title>
</head>
<body>
    input:<input id="id" type="text" name="" value="" /><br></br>
    output:<input id="op" type="text" name="" value="" /><br></br>
    <input type="button" value="save" name="save" id="save"/>
   <div id="output"></div>
</body>

test.jsp :

<%@page contentType="text/html" pageEncoding="UTF-8"%>
<% String id = request.getParameter("input");
String output = request.getParameter("output");
%> 
<%=id%> is stored <br/>output is:<%=output%>

The first screen shows this:

And after inserting data and pressing save button it shows this:

回答2:

Sorry, I would not post this as an answer but I have to because I do not have the privilege in my account to post it as a comment but I as sure your confuse what to be click. The #call refers to the element div not the actual button but it will work either. Can you please post your test.jsp so that we can see the code because I suspect that there cause the problem.

I think you had edited your question so I think my answer is a wasteful.