Why Doesn't This Work?!

D_tunisia

Can anyone help please? I have the code below:

<?php 

session_start();

require_once ('../../mysql_connect.php'); // Connect to the database.

?>

<div id="Layer1" style="position:absolute; left:12px; top:14px; width:669px; height:26px; z-index:1"><span class=\"style1\"><?php virtual('../includes/header.php'); ?></span></div>
<div id="Layer3" style="position:absolute; left:80px; top:86px; width:521px; height:172px; z-index:3">

<?php

$result = mysql_query("SELECT * FROM links WHERE $link_id = $id");

while($myrow = mysql_fetch_array($result)) {

?>

<p>&nbsp;</p>
<p align="center"><span style="You are now being sent to: <?php echo "$myrow[link]"; ?></p>
<meta http-equiv="refresh" content="2;url="<?php echo "$myrow[link]"; ?>">

</div>

<?php

}

?>

I get this error:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/jclickth/public_html/links/goto_link.php on line 0

Houdini

I do not see where you connected to your database server and also where you selected a database as in

$connect = mysql_connect($host,$user,$pass) 
    or die("Could not connect to server ".mysql_error()); 
$db = mysql_select_db($database,$connect) 
    or die("Could not select the database $database ".mysql_error());

That would be one good reason.

D_tunisia

Houdini wrote:
I do not see where you connected to your database server and also where you selected a database as in
$connect = mysql_connect($host,$user,$pass) 
    or die("Could not connect to server ".mysql_error()); 
$db = mysql_select_db($database,$connect) 
    or die("Could not select the database $database ".mysql_error());
That would be one good reason.

I connect in line 3 of the code.

require_once ('../../mysql_connect.php'); // Connect to the database.

Houdini

Are you echoing errors from that file (mysql_connect.php) because if not you might want to because it looks like a connection problem.

Roger_Ramjet

Der :quiet:

$result = mysql_query("SELECT * FROM links WHERE $link_id = $id");

thorpe

Also... Where is $id being defined? Your query is failing, hence mysql_query() is returning false and not a valid MySQL result resource. You should always check your query actually succeeded before trying to use the result. eg...

$result = mysql_query($sql);
if ($result) {
  while($row = "blah") {
    // ...
  }
} else {
  echo mysql_error();
}

Blulagoon

$result = mysql_query("SELECT * FROM links WHERE $link_id = $id");

Try formulating your query like this so you can see what is going on, then of course revert back to your previous style once it is all working.


$sql = "SELECT * FROM links WHERE $link_id = $id";
echo $sql;

Like Thorpe, I don't think that a value for $id is showing up in the query, echoing the query string to screen will confirm this or not.

HTH - Blu.

Roger_Ramjet

The problem, people, is that there is no column defined in the where clause: $link_id is not a valid column name, it is an undefined variable, so of course the query fails; probably with an error message something like this

you have an error in your sytax, check the manual for the correct syntax to use near

'WHERE = 4'

or whatever $id happens to be. See, no column name to compare it to.

thorpe

Notice I said 'Also...' at the start of my post? I saw $link_id highlighted in your reply.

Roger_Ramjet

Sorry.