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:[EMAIL PROTECTED]]
Sent: Friday, January 26, 2001 1:21 AM
To: [EMAIL PROTECTED]
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
[EMAIL PROTECTED]
"There is nothing more rewarding than reaching the goal you set for
yourself"
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