tehnightmare
New Member
I'm trying to learn and I'm stuck. I don't understand why this doesn't work. If I just leave the include and remove the function call and don't wrap the database connection in a function it works properly.What is it that I'm missing here?Error Message:\[quote\] Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in /home3/badamsne/public_html/views/dogs.php on line 24 Database query failed: \[/quote\]Web page code:\[code\] <?php include("../model/db_conn.php"); db_conn();?><!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"><html><head> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8"> <title></title></head><body> <?php // 3. Perform database query $result = mysql_query("SELECT * FROM dogs", $connection); if(!$result) { die("Database query failed: " . mysql_error()); } // 4. Use returned data while ($row = mysql_fetch_array($result)) { echo $row[0]." ".$row[1]."<br />"; } ?></body></html><?php// 5. Close connection mysql_close($connection);?>\[/code\]PHP Function in separate file:\[code\]<?phpfunction db_conn() { // 1. Create database connection $connection = mysql_connect("localhost","website_admin","p@ssw0rd"); if(!$connection) { die("Database connection failed: " . mysql_error()); } // 2. Select a database to use $db_select = mysql_select_db("website_db", $connection); if(!$db_select) { die("Database selection failed: " . mysql_error()); }}?>\[/code\]Thanks!Tom