$db = mysql_connect("localhost","vlad","123");
mysql_select_db("first",$db);
$result = mysql_query("SELECT * FROM first",$db);
while ($myrow = mysql_fetch_array($result)) {
  echo $myrow["name"];
}

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in Z:\home\localhost\www\myscl.php on line 13

<?php 
$db = mysql_connect("localhost","vlad","123");
mysql_select_db("first",$db);
$result=mysql_query("SELECT * FROM first",$db);
$myrow = mysql_fetch_array($result);
echo $myrow["name"] 

?>

В чем проблема ?