[ArgonX]

Hello,

I have some problem with php and getting this error

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource
in /home/xxx/public_html/xxxex/include/functions.php on line 240

The Line 240 is :

if (!$Row=mysql_fetch_array($Result))


DOES anyone know what is error here...

full code... -->

// Legge da file le informazioni riguardanti il file specificato in upload
//
function get_file_description($filename, $max_caracters = 0, $replacecharacters = 1)
{
global $DBName, $DBLink;

$Query= "SELECT * from fileinfo where filename='$filename'";
$Result= @mysql_db_query($DBName,$Query,$DBLink);
if (!$Row=mysql_fetch_array($Result)) --- line 240
{
return array('','','');
} else
{

[Keith]

I don't think the exclamation point can be there.

[v1brazy]

Try

if (!($Row=mysql_fetch_array($Result)))

peace!!

[ArgonX]

I try both and it didnt work. I still have thiss error. The script is working , i just have this error that really bother me.

[darkfreak]

Originally Posted by ArgonX

$Query= "SELECT * from fileinfo where filename='$filename'";
$Result= @mysql_db_query($DBName,$Query,$DBLink);

You could probably get some more info on errors if didn't use the @ before your code actually works.

From PHP manual:

resource mysql_db_query ( string database, string query [, resource link_identifier ] )
Note:
This function has been deprecated since PHP 4.0.6. Do not use this function. Use mysql_select_db() and mysql_query() instead.