dyea.marcello

I am running a PHP script, and keep getting errors like: *Notice: Undefined variable: result2 in D:\xampp-win32\htdocs\sites\bookstore\result_page.php on line 200 *warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given in D:\xampp-win32\htdocs\sites\bookstore\result_page.php on line 200 Line 200 looks like this: `<?php while($row = mysqli_fetch_array($result2))` What do they mean?

I am trying to display a list of my book However I can't select the database: this is a piece of my code: <?php $mysql_hostname = "localhost"; $mysql_user = ""; $mysql_password = ""; $mysql_database = "mag_rocket"; $prefix = ""; $bd = mysql_connect($mysql_hostname) or die("Could not connect database"); mysql_select_db($mysql_database, $bd) or …

I am trying to display a list of the book which category but is not running. this is a piece of my code: <?php require_once('auth.php'); $con=mysqli_connect("localhost","mag_install","mag123","mag_rocket"); // Check connection if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } $result = mysqli_query($con,"SELECT * FROM category"); ?> <!DOCTYPE …