PHP调用未定义的函数mysql_connect

Im getting an error "PHP call to undefined function" i don't know what the cause of this error can someone give me a clue how to solve this? im new to html and css.

here is the error. enter image description here

here is my sql enter image description here

here is my php code.

<?php
//for connecting db
include('connect.php');
if (!isset($_FILES['image']['tmp_name'])) {
echo "";
}
else
{
$file=$_FILES['image']['tmp_name'];
$image= addslashes(file_get_contents($_FILES['image']['tmp_name']));
$image_name= addslashes($_FILES['image']['name']);
move_uploaded_file($_FILES["image"]["tmp_name"],"gallery/" . $_FILES["image"]["name"]);
$photo="gallery/" . $_FILES["image"]["name"];

$query = "insert into images (photo)VALUES('$photo')";
$result = mysql_query($query); 

echo '<script type="text/javascript">alert("image successfully uploaded ");window.location=\'index.php\';</script>';
}
?>
<!DOCTYPE html>
<html>
    <head>
        <link href="css/style.css" rel="stylesheet" />
        <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.1/jquery.min.js"></script>
        <script src="js/slider.js"></script>
        <script>
        $(document).ready(function () {
        $('.flexslider').flexslider({
        animation: 'fade',
        controlsContainer: '.flexslider'
        });
        });
        </script>
    </head>
    <body>
    <div class="container">
    <form class="form" action="" method="POST" enctype="multipart/form-data">
        <div class="image">
            <p>Upload images and try your self </p>
        <div class="col-sm-4">
              <input class="form-control" id="image" name="image" type="file" onchange='AlertFilesize();'/>
              <input type="submit" value="image"/>
            </div>
        </div>
        </form>
        <div class="flexslider">
            <ul class="slides">
                <?php
                    // Creating query to fetch images from database.
                    $query = "select * from images order by Id desc limit 5";
                    $result = mysql_query($query);
                    while($r = mysql_fetch_array($result)){ 
                ?>
                    <li>
                    <img src="<?php echo $r['photo'];?>" width="400px" height="300px"/>
                    </li>
                <?php 
                } 
                ?>
            </ul>
        </div>
    </div>
    </body>
</html>

here is my sql code.

<?php
// hostname or ip of server
$servername='localhost';
// username and password to log onto db server
$dbusername='root';
$dbpassword='';
// name of database
$dbname='pegasus';

////////////// Do not  edit below/////////
$link=mysql_connect("$servername","$dbusername","$dbpassword");
if(!$link){
    die("Could not connect to MySQL");
    }
mysql_select_db("$dbname",$link) or die ("could not open db".mysql_error());

?>

The most up to date and best practice for using MySQL with PHP is the object-oriented solution. Here is an example:

define("HOST", "localhost");
define("USER", "username ");
define("PASS", "password");
define("DATABASE", "pegasus");

$mysqli = new mysqli(HOST, USER, PASS, DATABASE);
$q = "UPDATE products SET amount_sold = 6";
$mysqli->query($q);

Here is more info on this coding practice:

http://codular.com/php-mysqli

Hope this helps.