将数据发送到DB并通过AJAX返回

Question

The following code works properly and I am using it 'cause I need to update some notes in DB without refreshing a page. But what have to do when I need in the same event get some data back?

index.php where change should occur

<?php
$s = $GET['...']
$p = $GET['...']
print '
 <form id="notesForm">
  <textarea name="c">'.$note.'</textarea>
  <input type="text" value="LAST_CHANGE_NEED_GIVE_BACK_FROM_DB">
  <input type="hidden" name="s" value="'.$s.'">
  <input type="hidden" name="p" value="'.$p.'">
  <input type="submit" value="save note">
  <div id="response"></div>
</form>
';
?>

script which is posting data to live_notes.php and probably should be changed because of the goal

$(document).ready(function(){
    $('#notesForm').submit(function(){
        event.preventDefault();
        $('#response').html("<b>saving...</b>");
        $.ajax({
            type: 'POST',
            url: 'notes/live_notes.php',
            data: $(this).serialize()
        })
        .done(function(data){
            $('#response').html(data);
        })
        .fail(function() {
            alert("bad luck");
        });
        return false;
    });
});

notes/live_notes.php

<?php
$s = $_POST['s'];
$p = $_POST['p'];
$c = $_POST['c'];

// connecting DB

mysql_query("
UPDATE `poznamky`
SET
    last_change = now(),
    page = '$p',
    content = '$c'
WHERE
    page = '$p';
");

any idea?

Answer 1

<?php
$s = $_POST['s'];
$p = $_POST['p'];
$c = $_POST['c'];

// connecting DB

$blnSuccess = mysql_query("
UPDATE `poznamky`
SET
    last_change = now(),
    page = '$p',
    content = '$c'
WHERE
    page = '$p';
");

if($blnSuccess){
echo "success";
}
else {
echo "not success!";
}

in your jquery :

.done(function(data){
            $('#response').html(data); 
}

data have success or not success! which return as ajax response.

Update
If you want to return some data from database then:

$result = mysql_query("
   select * 
      from `poznamky`
");

while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
    echo $row['content'];
}

Then it will return all content field from poznamky table into data in ajax.