Need help to get rid of this - Warning:

[kprice]

This error appears, and I don't know how to fix it. I'd appreciate any assitance.

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource

Here is the code:

<?php

if($_COOKIE['phpbb3_7rhw4_u'] == "1"){
//The user isnt logged in, display the form

?>


<div class="formcontainer">
			<form action="../community/ucp.php?mode=login" method="POST" target="_parent">
			<p style="margin-top:0px" class="login">Username:</p>
			<input type="text" name="username" maxlength="30" size="15" value="" /> 
			<p style="margin-top:0px" class="login">Password:</p>
			<input type="password" name="password" maxlength="30" size="15" value="" /> </p>
			<p style="margin-top:0px" class="login"><input type="checkbox" name="autologin"  />
			Remember me&nbsp;&nbsp;&nbsp;&nbsp;</p>
			<input type="hidden" name="redirect" value="../" target="_parent"/>
			<p class="submit" style="margin-top:0px; margin-left:30px"> <input type="submit" name="login" value="login" /></p>

		<p style="margin:0px 0 0 16px" class="login"><a href="/community/ucp.php?mode=sendpassword" target="_parent">Forgot Password?</a></p>
		</form>
		</div>


<?php

}
else{
//The user is logged in, display some of the users info

   $userid = $_COOKIE['phpbb3_7rhw4_u']; // userid cookie
   $sid = $_COOKIE['phpbb3_7rhw4_sid']; // user session cookie

   $host = 'mysqlhost' ; // Db Host
   $db = 'xxxxxxx'; // Db Name
   $pass = 'xxxxxx'; // Db Pass
   $username = 'xxxxx'; // Db Username
   $link_id = mysql_connect($host,$username,$pass);
   if (!mysql_select_db($db)) die(mysql_error());

   $query = mysql_query("SELECT * FROM phpbb_users WHERE user_id = ". $userid);

   $data = mysql_fetch_array($query);



?>

site address: http://www.bornstorytellers.net

Many thanks in advance

Kevin

[laserlight]

That warning is due to one or more of these:

• Failed to connect to database server.

• Failed to select database.

• Failed to execute the corresponding SQL statement.

Because you have checks for the first two, the problem is most likely due to an invalid SQL statement.

[kprice]

Sorry I wasn't thinking clearly when I put this up. The error is reported to be in this line:

$data = mysql_fetch_array($query);

Therefore, I believe the problem is in my $query.

The trouble is, I don't know what is at fault in the $query.

This fault does not occur on my browsers, only on others going to the site to login.

All login operations function correctly. Visitors can sometimes remove the fault by refreshing the page, but this doesn't work on all browsers.

Let me know what you think. Thanks,

Kevin

[laserlight]

This fault does not occur on my browsers, only on others going to the site to login.

All login operations function correctly. Visitors can sometimes remove the fault by refreshing the page, but this doesn't work on all browsers.

That is unusual, since PHP operates on the server side.

I notice that $userid comes from a cookie, and yet you are not properly checking that the cookie value exists before using it. Use [man]isset[/man] or [man]empty[/man] to check that $_COOKIE['phpbb3_7rhw4_u'] exists, and only then assign it to $userid. When you do assign it, cast it to int so as to avoid SQL injection.

[kprice]

Thanks Laserlight,

I think I'm following you here...

Because the cookie likely exists in my browser history, the fault doesn't show up. But on a visitor's browser who has not previously logged into the system, the cookie would not exist, or on a browser who does not have cookies enabled.

Therefore, line 1 of the code would generate a non valid response...although I don't understand why it would manifest itself in line 64 (the line it's reported in).

However, not being a coder, I'm not sure how I would use the solution you propose, can you help me out with this?

Thanks,

Kevin

[laserlight]

Therefore, line 1 of the code would generate a non valid response...although I don't understand why it would manifest itself in line 64 (the line it's reported in).

One possibility is that your SQL query becomes:

SELECT * FROM phpbb_users WHERE user_id =

This is syntactically incorrect since there is nothing to compare user_id to, thus it will lead to the "supplied argument is not a valid MySQL result resource" warning that you have.

I think that a solution is to change:

if($_COOKIE['phpbb3_7rhw4_u'] == "1"){

to:

if(empty($_COOKIE['phpbb3_7rhw4_u'])){

and change:

$userid = $_COOKIE['phpbb3_7rhw4_u']; // userid cookie

to:

$userid = (int)$_COOKIE['phpbb3_7rhw4_u']; // userid cookie

[kprice]

Thank you Laserlight, I'll try that and let you know how I fare.

Much appreciated,

Kevin

[kprice]

Laserlight,

Your solution appears to have worked.

Thanks for all your help, I really do appreciate it.

Cheers,

Kevin

[laserlight]

My pleasure 🙂
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