RE: [PHP] Urgent help needed, sound scary when anyone did that on title :-)

you know Jacky, there's another, less cool and less reliable way to get the last inserted id: SELECT id FROM table ORDER BY id DESC; it will sort them all giving you the biggest id *number* (not what mysql keeps) and you can keep it for as many milliseconds your script will run more... So if nothing else works for you - try this... (depends on how secure & stable you want your application to be, of course...) Cheers, Maxim Maletsky -----Original Message----- From: Jacky@lilst [mailto:jacky@activelifestyle.com] Sent: Friday, January 26, 2001 1:21 AM To: php-general@lists.php.net Subject: [PHP] Urgent help needed, sound scary when anyone did that on title :-) Hi people, I got here the syntax that is suppose to get the id from the "just inserted" record and store it in value, did not work so far and I cannot see what is wrong in there, can anyone give me a hint what is wrong here? ( And the reason I did not use mysql_insert_id here is because the ID field at my tables are all BIGINT so mysql_insert_id won't work, so I have to use LAST_INSERT_ID() instead). By the way, the error after the page is executed keep saying that "Mysql warning : 0 is not Mysql index" ( and point to the line "$FirstLast = mysql_result($resultlast,0,0);"). And I did echo for the value of $FirstLast, it showed that there is no value in there. Sniplet is like this: ( I tried to keep this down as much as try to give most detail at the same time, so apologize for too long sniplet). $sql1 = "insert into firsttable (firstname, lastname) values('Jack','Chan')"; $resultsql1 = mysql_query($sql1); $sqlLastID = "select LAST_INSERT_ID() from firsttable"; $resultlast = mysql_query($sqlLastID); $FirstLast = mysql_result($resultlast,0,0); $sql2 = "insert into secondtable (FirsttableID,secfirstname, seclastname) values('$FirstLast','Jacky','Chany')"; $resultsql2 = mysql_query($sql2); $sqlLastIDsec = "select LAST_INSERT_ID() from secondtable"; $resultlast2 = mysql_query($sqlLastIDsec); $secondLast = mysql_result($resultlast2,0,0); $sql3 = "insert into Thirdtable (SecondTableID,FirsttableID,Thirdfirstname, Thirdlastname) values('$secondLast','$FirstLast','Steve','Chan')"; $resultsql3 = mysql_query($sql3); $sqlLastIDthird = "select LAST_INSERT_ID() from Thirdtable"; $resultlast3 = mysql_query($sqlLastIDthird); $ThirdLast = mysql_result($resultlast3,0,0); ****************** what have I done wrong? Please enlighten me here Thanks Jack jacky@activelifestyle.com "There is nothing more rewarding than reaching the goal you set for yourself"