jquery ajax calling servlet method and return json data

Question

I am new to Ajax programming I am using Jquery ajax my requirement to to fetch the data into server and display as table format.In that location must display radio button type. how to call servlet specific method how to return data.i am ready to use json.can any one help me how to call a method how to return a data.and suggestions require to solve the problem

thanks in advance.

$('#ajaxbutton').on('click',function(){
                $.ajax({
                    type:"post",
                    url:"Db2",
                    data:{"labid",100},
                    sucess:function(){
                        alert("sucess");
                    }

                });
            });

in servlet

public class Db2 extends HttpServlet {
    public void doGet(HttpServletRequest request, HttpServletResponse response)
            throws MalformedURLException, IOException {
        doProcess(request, response);
    }

    public void doPost(HttpServletRequest request, HttpServletResponse response)
            throws MalformedURLException, IOException {
        doProcess(request, response);
    }

    public void doProcess(HttpServletRequest request,
            HttpServletResponse response) throws MalformedURLException,
            IOException {
        Connection con;
        PreparedStatement ps, ps1;
        PrintWriter out = response.getWriter();

        try {

            Class.forName("sun.jdbc.odbc.JdbcOdbcDriver");
            con = DriverManager.getConnection("jdbc:odbc:oracle1", "system",
                    "sarath");
            ps = con.prepareStatement("select trackid,location,to_char(mydate,'dd-mon-yyyy') from information where labid=100");
            ResultSet rs = ps.executeQuery();

            while (rs.next()) {
                String location = rs.getString(2);
                String track = rs.getString(1);
                String myDate = rs.getString(3);
            }

        } catch (Exception e) {
            out.println(e);
        }

    }
}// end of ServletB

SpringLearner · Accepted Answer · 2013-11-02 06:27:32Z

2

how to call servlet specific method In your ajax you are writing type=post so in the servlet doPost() will be called.if it would have been type=get then servlet doGet() method will be called.So you should write the database retrieve part in the specific method.

Now suppose you want location,track and mydate then try this way In the servlet

PrintWriter out.response.getWriter();
out.println(location);
out.println(track);
out.println(mydate);

In the ajax success part do this way

success: function(data, textStatus, jqXHR){
                       alert(data);

edited Nov 2, 2013 at 6:27

answered Nov 2, 2013 at 6:21

SpringLearner

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7 Comments

Sarath Babu Nuthimadugu Over a year ago

can i use any other method in the type insted of get/post and can i write return "json"; in method. i am using only servlet for entire application

SpringLearner Over a year ago

see in servlet doPost or doGet will be called as per the form method(i.e get or post).

Sarath Babu Nuthimadugu Over a year ago

thanks it is working fine but i have doubt if i use so many data base quiries in one servlet how to differentiate only that method . generally we write public String method(){ return "jsonstring";} but here not possible to write return statement but unnecessarily calling entire servlet

SpringLearner Over a year ago

@sarath sorry could not understand

SpringLearner Over a year ago

@sarath you can write out.println(method());

|

Dawson Toth · Accepted Answer · 2014-12-12 05:50:38Z

0

PrintWriter out = response.getWriter();
StringBuffer res = new StringBuffer();
while (rs.next()) {
    String location = rs.getString(2);
    String track = rs.getString(1);
    String myDate = rs.getString(3);
    res.append("{");
    res.append("'location':");
    res.append(location);
    res.append(",");
    res.append("'track':");
    res.append(track);
    res.append(",");
    res.append("'myDate ':");
    res.append("myDate ");
    res.append("}");
}
out.println(res.toString());

must be json-string and the ajax can use eval("data = "+r_data+";");

edited Dec 12, 2014 at 5:50

Dawson Toth

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answered Nov 2, 2013 at 6:48

heqing

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6 Comments

Sarath Babu Nuthimadugu Over a year ago

it is returning json string i converted JSON.parse() now i want display a table format first record must be input type radio

heqing Over a year ago

display first record？ loop json datas to get first record...is it?(是这个意思吗?)

Sarath Babu Nuthimadugu Over a year ago

{'location':hyderbad,'track':12345,'myDate ':15-dec-2013}{'location':sydny,'track':2322,'myDate ':21-jan-2014} retuning data in this format it is showing error parse to json

heqing Over a year ago

{'location':'hyderbad','track':'12345','myDate ':'15-dec-2013'}{'location':'sydny','track':'2322','myDate ':'21-jan-2014'} it is json data style

Sarath Babu Nuthimadugu Over a year ago

it needs "," from one object another object

|

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