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 teh value of $FirstLast, it showed that there is no value in there.
Sniplet is like this:
$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
[EMAIL PROTECTED]
"There is nothing more rewarding than reaching the goal you set for yourself"
