Inserting checkbox values into MySQL Database with PHP

Checkboxes are only posted when they are ticked. So if the checkbox is not ticked, is will not appear in $_POST. Also you should generally not give any values to checkboxes. Instead use names to distinguish between them.

In the database I usually represent checkboxes with tinyints and store 1 for checked and 0 for unchecked.

// If your checkbox name is foo, this will convert it
// into a value that can be stored in the database
$foo = isset($_POST['foo']) ? 1 : 0;

Also note that html requires ids to be unique. So you can't have multiple elements with the same id. And you should sanitize your inputs to prevent sql injection. Use mysql_real_escape_string() on user inputs that go in the database.

Update

The main issue is that the query is missing a ')' at the last line. The query should look like

$query = "INSERT INTO markers (ciudad,
                        zona,address,name,
                        telefono,email,piso,
                        tipo,erasmus,nhabitaciones,
                        plazas,equipHabita,nbanos,
                        salon,cocina,electrodomesticos,
                        garaje,internet,calefaccion,
                        sexo,precio,superficie,otros,
                        fecha,lat,lng)
    VALUES ('{$_POST['ciudad']}','{$_POST['zona']}',
            '{$_POST['address']}','{$_POST['name']}','{$_POST['telefono']}',
            '{$_POST['email']}','{$_POST['piso']}','{$_POST['tipo']}',
            '{$_POST['erasmus']}','{$_POST['nhabitaciones']}',
            '{$_POST['plazas']}','{$equipHabitaF}',
            '{$_POST['nbanos']}','{$equipSalonF}',
            '{$equipCocinaF}','{$equipElectroF}','{$_POST['garaje']}',
            '{$_POST['internet']}','{$_POST['calefaccion']}',
            '{$_POST['sexo']}','{$_POST['precio']}',
            '{$_POST['superficie']}','{$_POST['otrosF']}',
            '{$_POST['fecha']}','{$_POST['lat']}',
            '{$_POST['lng']}')";
mysql_query($query, $link);

Note the closing ')' on the last line of the query. Also note that creating the query like this, in a variable, allows you to output the created query so you can see what exactly is sent to MySQL and you also can test your query in a different environment (ie in phpmyadmin or any other database administration tool).

Name your form in html as -

<form id="prices_form" name="pricesForm" method="post" action=""> 

Try JSON.stringify() data before sending with the AJAX like below -

var data = JSON.stringify({ 
    action: 'ajaxsavePrice', 
    location1: $('#location1').val(), 
    location2: $('#location2').val(), 
    price: $('#price').val() 
});

And then replace your ajax call on form submit as below-

$('form.pricesForm').on('submit', function(e){
  e.preventDefault();
  $.ajax({
   method: 'POST',
   dataType: 'json',
   url: ajax_savePrice_object.ajaxurl, // also check this if it returns the correct url
   data: data,
   success: function(res){
     $('#prices_form').hide();
   }
  });

});

Hope this helps.

Let me clearify you in an easy way...

This will be your controller

class Site extends CI_Controller
{
function index()
{
$this->load->view('form.php');// loading form view
}

function insert_to_db()
{
$this->load->model('site_model');
$this->site_model->insert_to_db();
$this->load->view('success');//loading success view
}
}

This will be your views. Befor creating views go to autoload.php and autoload url helper and database class. In your config.php set config['base_url'] = "path to your site"; form.php

<form action="<?php echo base_url();?>index.php/site/insert_into_db" method="post">
Field 1 = <input type = 'text' name='f1'>
Field 2 = <input type = 'text' name='f2'>
Field 3 = <input type = 'text' name='f3'>
<input type='submit'>
</form>

success.php

<b>Your data has been inserted!!!</b>

In your model you need to pull those datas of form and insert into db this way

site_model.php

class Site_model extends CI_Model
{
function insert_into_db()
{
$f1 = $_POST['f1'];
$f2 = $_POST['f2'];
$f3 = $_POST['f3'];
$this->db->query("INSERT INTO tbl_name VALUES('$f1','$f2','$f3')");
}
}

In this case your database has three fields... modify query as per your requirements