Not sure why my codes stopped working

rdkd1970

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;
  }
}
  }

?>
<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>

NogDog

I would guess that the mysql_connect() in your connection include file is failing. (Note that testing for a failure of include() won't tell you the mysql_connect() failed, unless you explicitly [man]return[/man] false in that include file in such a case.)

rdkd1970

I thought the or die ("could not connect to db") was to let me know it was not connecting. However I was doing fine with the connection as I had several updates to the db??????

NogDog

The "or die()" in your code is only telling you if the include() worked, which can work just fine even if the file it includes has a problem. You would need to add the "or die()" to the actual mysql_connect() command within that include file if that's what you want to happen (or probably better would be to throw an exception and try/catch it in the main script, but that's another story).

In any case, the error messages are indicating that there is no database connection, so it is trying to automatically connect using the current user ID with no password, and is failing (no big surprise).

rdkd1970

thanks I just deleted that or die put the ; after the connection and also deleted some information in the db connection and it worked fine thanks.