Someone jump in here and correct me if I'm wrong, but by the time you
get around to executing the second query, LAST_INSERT_ID() from MySQL
isn't necessarily going to be the desired value, because another record
may well have been inserted in that time?
On 24 Jan 2001 18:25:27 +0900, Maxim Maletsky wrote:
> for example:
>
> $SQL = "INSERT INTO users SET name='Maxim', surname='Maletsky'";
>
> now you have to insert into another table where you need to relate that user
> to the entry:
>
> $SQL2 = "INSERT INTO questions SET question='how did you sleep?',
> made_by=LAST_INSERT_ID()";
>
> LAST_INSERT_ID() will be here equal to the auto_incremented id of the first
> $SQL statement.
>
> Hope this helps,
> Maxim Maletsky
>
>
> -----Original Message-----
> From: Jacky@lilst [mailto:[EMAIL PROTECTED]]
> Sent: Thursday, January 25, 2001 7:16 AM
> To: [EMAIL PROTECTED]
> Subject: [PHP] last_insert_id function
>
>
> I got this quote right out of the php manual. My Id field happen to be type
> BIGINT as it said so I tried using LAST_INSERT_ID(); and turn out to be
> error said "unidentified function". Any clue?
> *******************************************
> 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().
> ********************************************
> Jack
> [EMAIL PROTECTED]
> "There is nothing more rewarding than reaching the goal you set for
> yourself"
>
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