When my ajax is processed
$.post("main.php",
{data: $(this).text()},
function(data) {
//alert("Data saved.");
$('#demo').html(data);
});
instead of getting what I specified in the below query
if (isset($_POST['data'])){
$data = $_POST['data'];
$query = mysql_query("SELECT * FROM tempusers
WHERE 'firstname' LIKE '%$data%'
OR 'lastname' LIKE '%$data%'
OR 'title' LIKE '%$data%'");
while($row=mysql_fetch_assoc($query)){
$firstname=$row['firstname'];
$lastname=$row['lastname'];
$grade=$row['grade'];
echo $grade;
Instead I get all the elements in my page returned. So in other words in the div below
<div id="demo"></div>
I'm returning my page. So it shows as a website within a website. It won't even show echo $grade;
I thought it was my query acting up, but I tried commenting out the query and just echo $data
see below
if (isset($_POST['data'])){
$data = $_POST['data'];
echo $data;
And doing so gave me the same result as stated previously "echoing out the website within the website" and also echoed the $data I wanted. I would be overjoyed to understand how this happened as well as what I can do to fix this problem and get my isset function/ajax corrected. Any other tips would be greatly appreciated. I know I need to change the query to PDO and plan on it.