rdkd1970 Posted May 3, 2011 Share Posted May 3, 2011 I am not sure what is going wrong but I have not changed any of my codes and now I am getting this error message on a simple form that I am testing. Any idea what has happened. I have had several successful tries that have updated to the db. Warning: mysql_query() [function.mysql-query]: Access denied for user 'ebermy5'@'localhost' (using password: NO) in /home/ebermy5/public_html/form.php on line 20 Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in /home/ebermy5/public_html/form.php on line 20 Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/ebermy5/public_html/form.php on line 21 Warning: mysql_query() [function.mysql-query]: Access denied for user 'ebermy5'@'localhost' (using password: NO) in /home/ebermy5/public_html/form.php on line 29 Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in /home/ebermy5/public_html/form.php on line 29 This is the form <?php @include_once ("Connections/connect_to_mysql.php") or die('could not connect to database'); $err=''; if($_POST["submit"]){ // Validate form data if($_POST["firstname"]=='') $err.='Please enter First Name<br>'; if($_POST["email"]=='') $err.='Please enter Email<br>'; if($err==''){ // Check if there are duplicate entries in the 'contacts' table $results = mysql_query("SELECT id FROM `Members` WHERE firstname='".addslashes($_POST["firstname"])."' and Email='".addslashes($_POST["email"])."'"); if($row = mysql_fetch_array($results)){ $err.='Can not add duplicate entry<br>'; } else{ // adding new record to 'contacts' table mysql_query("INSERT INTO Members (firstname,lastname,country,Email) values ('".addslashes($_POST["firstname"])."','".addslashes($_POST["lastname"])."','".addslashes($_POST["country"])."','".addslashes($_POST["email"])."')"); // redirecting to success screen exit; } } } ?> The HTML below.. <html> <head> <title>Add New Contact</title> </head> <body> <h2>Register with us</h2> <?php echo $err==''?'''<p style="color:red;">'.$err.'</p>') ?> <form method="post" action="form.php"> <table border="0"> <tr> <td valign="middle">First Name:</td> <td><input type="text" name="firstname" size="30" value="<?php echo htmlspecialchars($firstname) ?>"></td> </tr> <tr> <td valign="middle">Last Name:</td> <td><input type="text" name="lastname" size="30" value="<?php echo htmlspecialchars($lastname) ?>"></td> </tr> <tr> <td valign="middle">Country:</td> <td><input type="text" name="country" size="30" value="<?php echo htmlspecialchars($country) ?>"></td> </tr> <tr> <td valign="middle">Email:</td> <td><input type="text" name="email" size="30" value="<?php echo htmlspecialchars($email) ?>"></td> </tr> </table><br> <input type="submit" name="submit" value=" Submit! "> </form> </body> </html> Link to comment https://forums.phpfreaks.com/topic/235395-not-sure-what-has-happened-with-codes/ Share on other sites More sharing options...
PFMaBiSmAd Posted May 3, 2011 Share Posted May 3, 2011 Your mysql_query() statement is attempting to create a connection to the database server, because your actual code that is supposed to be making that connection is failing. What is in your Connections/connect_to_mysql.php file? xxxxx out your username and password, but don't change anything else when you post the code. Post the opening php tag that is in that file as well. Link to comment https://forums.phpfreaks.com/topic/235395-not-sure-what-has-happened-with-codes/#findComment-1209726 Share on other sites More sharing options...
rdkd1970 Posted May 3, 2011 Author Share Posted May 3, 2011 <?php // Place db host name. Sometimes "localhost" but // sometimes looks like this: >> ???mysql??.someserver.net $db_host = "localhost"; // Place the username for the MySQL database here $db_username = "xxxxx"; // Place the password for the MySQL database here $db_pass = "xxxx"; // Place the name for the MySQL database here $db_name = "ebermy5_bermy"; // Place the port for the MySQL database here $mysql_connection_port = "3306"; set_time_limit(0); // Run the connection here $link = mysql_connect("$db_host","$db_username","$db_pass") or die ("could not connect to mysql"); $link = mysql_select_db("$db_name") or die ("Query died:"); $link = $mysql_connection_port = "3306"; ?> Link to comment https://forums.phpfreaks.com/topic/235395-not-sure-what-has-happened-with-codes/#findComment-1209738 Share on other sites More sharing options...
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