sql - PHP call to undefined function mysql_connect -
im getting error "php call undefined function" don't know cause of error can give me clue how solve this? im new html , css.
here error. 
here sql 
here 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"]; $query = "insert images (photo)values('$photo')"; $result = mysql_query($query); echo '<script type="text/javascript">alert("image 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 , try 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 fetch images database. $query = "select * images order 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 sql code.
<?php // hostname or ip of server $servername='localhost'; // username , password log onto db server $dbusername='root'; $dbpassword=''; // name of database $dbname='pegasus'; ////////////// not edit below///////// $link=mysql_connect("$servername","$dbusername","$dbpassword"); if(!$link){ die("could not connect mysql"); } mysql_select_db("$dbname",$link) or die ("could not open db".mysql_error()); ?>
the date , best practice using mysql php object-oriented solution. here 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 more info on coding practice:
hope helps.
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