title: Changing check box automatically changes the database record
body: <p>in my database I have the filed status which is boolean type, </p>
  
  <p>thank I have a connection to the database </p>
  
  <pre><code>    if (!$result = mysqli_query($con,"SELECT * FROM cursos"))
  {
      die("Error: " . mysqli_error($con));
  }
  </code></pre>
  
  <p>and displaying the status of this field with checkboxes,</p>
  
  <pre><code> &lt;?php if($row['status'] == 1) { 
      echo "&lt;td&gt;"."&lt;input type = 'checkbox' checked='checked' name ='complete' value= '1'/&gt;"."&lt;/td&gt;";
  }else
  {
      echo "&lt;td&gt;"."&lt;input type = 'checkbox'  name ='incomplete' value= '0'/&gt;"."&lt;/td&gt;";
  }
  ?&gt;
  </code></pre>
  
  <p>So what I need to do is, once I check or uncheck any of thoose check boxes, the database field to be canged automatically withoud the need of submit button. </p>
  
  <p><strong>Okay I did edited my code to what you suggested and it s working now, here is the new code:</strong></p>
  
  <p>Ajax Code</p>
  
  <pre><code>&lt;script&gt;
  $(document).ready(function(e) {
  $('.checkboxtest').change(function(){
      if( $('.checkboxtest').prop('checked') )
         {checkboxstatus = '1';}
         else
         {checkboxstatus = '0';}
  $.ajax({
          type: "POST",
          url: "checkboxtestbackend.php",
          data: {checkboxstatus: checkboxstatus},
          })
          .done(function(data, textStatus, jqXHR){alert(textStatus);})
          .fail(function(jqXHR, textStatus, errorThrown){alert(jqXHR+"--"+textStatus+"--"+errorThrown);});
  });//end change
  });//end ready
  &lt;/script&gt;
  </code></pre>
  
  <p>the html code is: </p>
  
  <pre><code>&lt;?php if($row['status'] == 1) { 
      echo "&lt;td&gt;"."&lt;input type = 'checkbox' class='checkboxtest' checked='checked' name ='complete' value= '1'/&gt;"."&lt;/td&gt;";
  }else
  {
      echo "&lt;td&gt;"."&lt;input type = 'checkbox' class='checkboxtest' name ='incomplete' value= '0'/&gt;"."&lt;/td&gt;";
  }
  ?&gt;
  </code></pre>
  
  <p>and the php backend code is: </p>
  
  <pre><code>&lt;?php
  $checkboxstatus = $_POST['checkboxstatus'];
  
  $host = "localhost";
  $user = "username";
  $password = "password";
  $dbname = "name";
  
  $cxn = mysqli_connect($host,$user,$password,$dbname);
  if (mysqli_connect_errno()) {echo "No connection" . mysqli_connect_error();}
  
  $query = " UPDATE cursos
             SET status = '$checkboxstatus'
             WHERE id = '12'";
  $result = mysqli_query($cxn, $query) or die ("could not query database 1");
  ?&gt;
  </code></pre>