The followoing error message: 
>Warning: Supplied argument is not a valid MySQL result resource

means that a php function hasn't been supplied witht the correct number 
of arguments.  Try the following:

$link = mysql_connect("host","yourusername","yourpassword") or 
die("Died at connection.");
table_name = "yourtable";
$sql = "SELECT * FROM yourtable";
$result = mysql_query ($result, connection);

$fields = mysql_num_fields($result);
$rows = mysql_num_rows($result);

$table = mysql_field_table($result, $i);

echo "Your '".$table."' table has ".$fields." fields and ".$rows." 
records <BR>"

- - EOP - -

//All I've done is to give mysql_query() two arguments
//Remover gaps between mysql functions anf their arguments (Which they 
//shouldn't have)

And given mysql_connect all 3 arguments it should have to make a 
successful connection to your database server. (Not sure wether you 
missed these out simply for clarity though)

Hope that all helps and makes sense!!


         "Believe nothing - consider everything"        
  Russ Michell
  Anglia Polytechnic University Webteam
  +44 (0)1223 363271 ext 2331

PHP Database Mailing List (http://www.php.net/)
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]

Reply via email to