$result = mysql_db_query($dname, $sql); $row = mysql_fetch_object($result);
echo "$row->field_name"; -- Justin Garrett "Melih Onvural" <[EMAIL PROTECTED]> wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... > i have a database and it randomly selects one piece of data from a table of > quotes, but it will only print the quote when I include the row that it's > picked with. I have my SELECT statement at > > $sql = "SELECT $dbrow FROM $dbtable ORDER BY RAND() LIMIT 1"; > > and then to display the result > > $result = mysql_db_query($dbname, $sql) > > > then in the code, > > print "<td> $result </td>"; > > what do i need to add? (the result is Resource id #3) > > thank you, > Melih > > -- PHP General 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]
