Question

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.

Answer 1

After you echo what you want, put an explicit exit to terminate any further script execution.

 echo $grade;
 exit;

If the grade echo is in a loop, put the exit after the loop termination.