note 83942 deleted from function.mysql-insert-id by cmb

Note Submitter: icethenet at NOSPAM dot hotmail dot com ---- Other methods seem to have problems with missing records in auto increment sometimes you will have records 1 2 5 6 most functions would return the value of 5 for next auto increment when indeed it would be 7. This is the only way I found to make this work so I can use my customer number and the record number to provide a truly unique customer number that is also useful. $next_increment = 0; $qShowStatus = "SHOW TABLE STATUS LIKE 'your_table'"; $qShowStatusResult = mysql_query($qShowStatus) or die(mysql_error()); $row = mysql_fetch_assoc($qShowStatusResult); $next_increment = $row['Auto_increment']; echo $next_increment; then you can do something like this echo $next_increment ."-". rand(); My first post: I hope this is useful to someone