On Thu, Oct 23, 2003 at 03:51:30PM -0700, Mike Alderson wrote: : : I am trying to gather information from one database, : process it, and insert it into a new one. When I run : this script I am getting the following error: : : Warning: mysql_fetch_array(): supplied argument is not : a valid MySQL result resource in : C:\web\aldersonfamily\convert.php on line 11 : : Here is my code: : : <? : : $connectionnew=mysql_connect("localhost","app","aardvark") : or die("Can't connect to database new"); : $dbnew=mysql_select_db("alderson_cms", : $connectionnew) or die("Can't select db : cms_alderson"); : : : $connectionold=mysql_connect("localhost","app","aardvark") : or die("Can't connect to database : $DBASEURL,$DBASENAME,$DBASEPASS"); : $db=mysql_select_db("aldersondb1", : $connectionold) or die("Can't select db aldersondb1"); : : $sql="SELECT * FROM users"; : $result=mysql_query($sql, $connectionold) or : die(mysql_error());
Are you sure your query succeeded before you entered your while loop? : While ($row=mysql_fetch_array($result)) : { : $id=$row['id']; : $fname=$row['fname']; : $lname=$row['lname']; : $password=$row['password']; : $username=$row['username']; : $security = crypt("user", 12); : $password = crypt($password, 12); : : $sql = "INSERT INTO cms_users (id, fname, : lname, username, email, password, security, origip) : VALUES (\"\", \"$fname\", \"$lname\",\"$username\", : \"\", \"$password\", \"$security\", \"\")"; : $result=mysql_query($sql, $connectionnew) or : die(mysql_error()); : } : : ?> -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php