Question

I don’t really know why I have this issue, I have looked over my code; and all looks well.

Here my “addtofavorites.php” page

<?php
$con = mysql_connect("localhost","root","student");


if (!$con);
{
die('Could not connect: ' . mysql_error());
}                   
    mysql_select_db("tvid", $con);

    $sql="INSERT INTO tv (userid, favorites) VALUES ('345','77');"
        if (!mysql_query($sql,$con));
        {
        die('Error: ' . mysql_error());
        }
    echo "Your Video was Added To Your Favorites";
    mysql_close($con);

?>

Here where I reference it on a Ajax listen event (listen for a button click)

$(document).ready(function () { 
$('#button_1').click(function(e){
    e.preventDefault();
    e.stopPropagation();
    favfunct();
});
});
function favfunct() {
        $.ajax({
 type: "POST",
 url: "js/addtofavorites.php",
 data: { "get" : "runfunction", "action" : "favorites1" },
 success: function (response) {
     alert ("successfully loaded")
 }    
});
}

Any error I have? or Advice You might be able to give?

Answer 1

Here, try this piece:

<?php
$con = mysql_connect("localhost","root","student");
if (!$con) {    //mis-placed semicolon here
    die('Could not connect: ' . mysql_error());
}                   
mysql_select_db("tvid", $con);
$sql="INSERT INTO tv (userid, favorites) VALUES ('345','77');"; //semicolon here
    if (!mysql_query($sql,$con));
    {
    die('Error: ' . mysql_error());
    }
echo "Your Video was Added To Your Favorites";
mysql_close($con);
?>

You current page will insert the values 345|77 repeatedly into the database.