mysql_fetch_array() gives error?

[Maurice]

Hi all,currently I'm trying to create a simple mySQL database. Using the tutorials iver here I've come pretty far. Now I have a question.The following code (can be find on http://w3schools.com/php/php_mysql_select.asp ) gives me the error that the mysql_fetch_array() doesn't excist or is invalid. In the $result there can be found:$result = mysql_query("SELECT * FROM leden");

while($row = mysql_fetch_array($result))  {  echo "<tr>";  echo "<td>" . $row['FirstName'] . "</td>";  echo "<td>" . $row['LastName'] . "</td>";  echo "</tr>";  }

Could somebody help me out on this, so I can read out the information from the 'leden' table? I think the error is the $row, so does the error from mysql thinks, but I don't know how to fix this. Complete code (copied from w3schools):

<?php$con = mysql_connect("localhost","******","******");if (!$con)  {  die('Could not connect: ' . mysql_error());  }mysql_select_db("leden__DB__test", $con);$result = mysql_query("SELECT * FROM leden");echo "<table border='1'><tr><th>Voornaam</th><th>Achternaam</th></tr>";while($row = mysql_fetch_array($result))  {  echo "<tr>";  echo "<td>" . $row['voornaam'] . "</td>";  echo "<td>" . $row['achternaam'] . "</td>";  echo "</tr>";  }echo "</table>";mysql_close($con);?>

and the table 'leden' looks like this:-------------------------------| voornaam | achternaam ||-------------|---------------||Maurice | Krielaart ||Admin | Admin |-------------------------------Hope somebody can help me getting the data from the table into the browser page.Thanks in advanceMaurice Krielaart

[Maurice]

Never mind, I inserted the incorrect name for the database in the code, however thanks all!