[FIXED] How to get jQuery .load(), .post() response in Spring MVC?

Issue

I am using jQuery in Spring MVC for validation purposes. Everything is OK related to validation, but the problem is that I cannot return any response from the target page to the page the request is sent. Here is the partial code of the view page ( from where the data is sent ):

$('#loginbtn').click( function() {
        var haserror = false;

        if( $('#username').val() == null || $.trim( $('#username').val() ) == '' ) {
            $('.username').removeClass('hide');
            haserror = true;
        } else {
            $('.username').addClass('hide');
        }

        if( $('#password').val() == null || $.trim( $('#password').val() ) == '' ) {
            $('.password').removeClass('hide');
            haserror = true;
        } else {
            $('.password').addClass('hide');
        }

        if( haserror == true ) {
            return;
        } else {
            //console.log('OK');
            $('#checklogin').load("insert", {'username':$('#username').val(), 'password':$('#password').val() }, function(d) {
                //alert(d);
            });
        }
    });

The above code is correctly validated data and sends the data to my target: insert

Here is the code written for insert:

@RequestMapping(value="/insert", method = RequestMethod.POST)
public String insert(Model model, @RequestParam(value="username", required=true) String username, @RequestParam(value="password", required=true) String password) throws SQLException {
    try {
        String query="INSERT INTO userlog(username, pword) VALUES('"+ username +"','"+ password +"')";
        Database.insert(query);
        return '1';
    } catch( SQLException E) {
        //E.printStackTrace();
        return '2';          
    }
}

Code for Database.insert(query):

package com.tng.test;

import java.sql.Connection;
import java.sql.DriverManager;
import java.sql.SQLException;
import java.sql.Statement;

public class Database {
    public static void insert( String query ) throws java.sql.SQLException {
        Statement stmt = null;
        Connection conn = null;
    
        try {
            Class.forName("com.mysql.jdbc.Driver");
    
            conn = DriverManager.getConnection("jdbc:mysql://localhost:3306/springtest","root","");  
            stmt= conn.createStatement();  
        
            stmt.execute(query);
        
            try {
                if(stmt != null) stmt.close();
                if(conn != null) conn.close();
            } catch (SQLException e) {
                e.printStackTrace();
            }
        } catch (ClassNotFoundException e1) {
            // TODO Auto-generated catch block
            e1.printStackTrace();
        }
    }
 }

I am trying to return '1' if data is successfully inserted and '2' if something goes wrong ( as usual I am doing in PHP project ). But it tries to redirect me to a page with the following url: base_url/1 and base_url/2, instead of sending back '1' and '2' as a response to my source page.

Can anyone tell me how I can fix the issue & send a response from the target page.

N.B: I am new in Spring MVC, above form & data insertion are just for testing, so no encryption of password is used.
Thanks

Solution

You need to add ResponseBody annotation in your post method.

org.springframework.web.bind.annotation.ResponseBody

@RequestMapping(value="/insert", method = RequestMethod.POST)
@ResponseBody
public String insert(Model model, @RequestParam(value="username", required=true) String username, @RequestParam(value="password", required=true) String password) throws SQLException {
    try {
        String query="INSERT INTO userlog(username, pword) VALUES('"+ username +"','"+ password +"')";
        Database.insert(query);
        return "1";
    } catch( SQLException E) {
        //E.printStackTrace();
        return "2";          
    }
}

Answered By - Rizwan M.Tuman