Warning: mysql_error() expects parameter 1 to be resource, string given in F:\ProductsQueries.php on line 43

[ziad]

Hi guys, I am trying to run the code below and I end up getting the error above, as am a newby in php, i just can't figure out where am going wrong, can anyone help me please

 

<?php
 require "db3.inc";

// 1.open the database connection - always use localhost
  $connection = mysql_connect($hostname, $username, $password);

 //2. Open the database connection and use the winestore database
  mysql_select_db($dbname, $connection);
 
  $query = "SELECT * FROM product";
  $ProductType = $_GET["ProductType"] ;
 
 
 //Create the Query
  $query = "SELECT  * FROM  product ";

            
  // ... then, if the user has specified a type ,
   // add the regionName as an AND clause ...
   
   if ($ProductType != 1)
 // Note the literals around $windType
   $query .= " WHERE Make = '$ProductType'";
   // ... and then complete the query.
   $query .= " ORDER BY Make ";
   

 //Get the recordset
 $result = mysql_query($query,$connection) OR die(mysql_error($query));
 
 

  $num_rows = mysql_num_rows($result);

  //Check some records have been found
   if ($num_rows < 1 )
   {
     echo "<strong> No Products found - make sure you entered the Product name correctly</strong>";
   }
   else
   //Format the information into a table using table/cell tags      
    while ($row = @ mysql_fetch_row($result))
    {
    // ... start a Table row...
    echo "\n<tr>";
    
// ... and print out each of the attributes in that row as
    // seperate TD (table Date)
    foreach($row as $data)
        print "\n\t<td> {$data} </td>";
        
    // finish the row
    print "\n<tr>";
    }
 // then finish the table
 print "\n</table>\n";


// (5) Close the connection
mysql_close($connection);

?>
</table>
</body>
</html>
 

 

[oaass]

$query is a string, $connection is a resource 
 
From the manual

string mysql_error ([ resource $link_identifier = NULL ] )

Try this instead

$result = mysql_query($query,$connection) OR die(mysql_error($connection));

I do strongly advice you to stop using the mysql extension and look at mysqli and/or PDO instead.. See my signature for the reason

[ziad]

Thanks for your suggestion oaass, Changing the $query string into the $connection Resource eliminated the error, but it created a new error saying "NO DATABASE SELECTED".

[DavidAM]

Please use CODE tags when posting code:

 

That error would indicate that you have not selected a database. Back at the beginning of the code you have:

 //2. Open the database connection and use the winestore database
  mysql_select_db($dbname, $connection);

add the or die(mysql_error()); to that statement to see why the database selection is failing. Either the database does not exist, or the user does not have access to it, or the variable is not defined, or something.