What wrong with this

Hi all

$query = "SELECT username, email, from customers where username = '$uname'";
$result = @($query);
$row = mysql_fetch_array($result,MYSQL_NUM);

I am receiving the following error with the above code any one have any ideas

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

echoed $query and the string is well formed.

any help appreciated

Mike

you have combined the mysql_fetch_assoc with the code for mysql_fetch_array

I would change to this

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

   //echo out data

}

//or

$row = mysql_fetch_assoc($result,MYSQL_NUM);  

//do something with data

}

Originally posted by mdesouza
$result = @($query);

Please delete this line!!! This kind of instruction can be used only if you are already a good coder...

Use instead this:
$result = mysql_query($query) or die("Error 1:".mysql_error()." Your query: ".$query."");

This gives you a usefull help string...

bye bye