Re: Urgent help needed, sound scary when anyone did that on title :-)

I don't think I can use mysql_insert_id() because the ID field in the tables are all BIGINT Auto_Increment and I saw that in the manual, it said mysql_insert_id will not work corerctly with this type of data. I hope I am wrong though. But if it is as the manual said, what else could I do? Direct quote from the manul at the part about mysql_insert_id() is here: ******************************************* mysql_insert_id() converts the return type of the native MySQL C API function mysql_insert_id() to a type of long. If your AUTO_INCREMENT column has a column type of BIGINT, the value returned by mysql_insert_id() will be incorrect. Instead, use the internal MySQL SQL function LAST_INSERT_ID(). ******************************************** cheers Jack jacky@activelifestyle.com "There is nothing more rewarding than reaching the goal you set for yourself" ----- Original Message ----- From: Joe Stump <jstump@zebulon.miester.org> To: Jacky@lilst <jacky@activelifestyle.com> Cc: <php-general@lists.php.net> Sent: Wednesday, January 24, 2001 9:25 PM Subject: Re: [PHP] Urgent help needed, sound scary when anyone did that on title :-) > remove the result - so just type $id = mysql_insert_id() and it should work > fine. > > --Joe > > > On Thu, Jan 25, 2001 at 10:21:17AM -0600, Jacky@lilst wrote: > > Hi people, > > I got here the syntax that is suppose to get the id from the "just inserted" record and store it in value, did not work so far and I cannot see what is wrong in there, can anyone give me a hint what is wrong here? ( And the reason I did not use mysql_insert_id here is because the ID field at my tables are all BIGINT so mysql_insert_id won't work, so I have to use LAST_INSERT_ID() instead). By the way, the error after the page is executed keep saying that "Mysql warning : 0 is not Mysql index" ( and point to the line "$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" > > -- > > Joe Stump, PHP Hacker > joestump98@yahoo.com > http://www.miester.org/ >