Right. I have successfully generate the variable I need with this code:
PHP:
PHP Code:
$query = sprintf("SELECT id FROM ibf_members WHERE name='%s'",
mysql_real_escape_string($username));
// Perform Query
$result = mysql_query($query);
// Check result
// This shows the actual query sent to MySQL, and the error. Useful for debugging.
if (!$result) {
$message = 'Invalid query: ' . mysql_error() . "\n";
$message .= 'Whole query: ' . $query;
die($message);
}
// Use result
// Attempting to print $result won't allow access to information in the resource
// One of the mysql result functions must be used
// See also mysql_result(), mysql_fetch_array(), mysql_fetch_row(), etc.
while ($row = mysql_fetch_assoc($result)) {
print $row['id'];
}
// Free the resources associated with the result set
// This is done automatically at the end of the script
mysql_free_result($result);
But I need to include the $row['id'] variable in a link.
The link should be in the following format:
http://forum.example.com/index.php?showuser=$row['id']
But I don't know how to write that in php.
I would also like 'View your forum CP' to be shown as the 'link'
I hope someone understands me, because even I'm not sure!