At 11:36 AM 10/13/2004, Paul DuBois wrote:
At 11:10 -0400 10/13/04, Michael Ragsdale wrote:
I've been using mysql_insert_id() with great success, but now I've got a
problem. I'm using UPDATE to, well, update a record in a database and
according to the docs...
mysql_insert_id() is updated after
- Original Message -
From: "Michael Ragsdale" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Wednesday, October 13, 2004 11:10 AM
Subject: mysql_insert_id() for UPDATE?
> I've been using mysql_insert_id() with great success, but now I've got a
> problem. I'm using UPDATE to, well, u
Didn't you have to specify the ID of the record in order to UPDATE it?
If not, and you updated several records, and assuming that the function
LAST_INSERT_ID worked as you had expected it to work, you would have still
only gotten one of the ID values for one of the updated records, right?
What
At 11:10 -0400 10/13/04, Michael Ragsdale wrote:
I've been using mysql_insert_id() with great success, but now I've
got a problem. I'm using UPDATE to, well, update a record in a
database and according to the docs...
mysql_insert_id() is updated after INSERT and UPDATE statements
that generate
Michael Villalba writes:
> mysql_insert_id returns the correct value to my application when I insert a
> single
> row using the INSERT...VALUES syntax. However, when I insert multiple rows,
> it always returns 0. This is inconsistent with the behavior of
> LAST_INSERT_ID,
> which returns the id
Hi Andrea,
> I am really new at PHP & MySQL, so please bear with me.
welcome to our happy band...
> I am using the mysql_insert_id function but continually receive an error,
> and hoping someone can point me in the right direction to resolve this.
>
> My error:
> Warning: Supplied argument is n
Hi Andrea,
> I am really new at PHP & MySQL, so please bear with me.
welcome to our happy band...
> I am using the mysql_insert_id function but continually receive an error,
> and hoping someone can point me in the right direction to resolve this.
>
> My error:
> Warning: Supplied argument is n
Hi.
You know, mysql_insert_id returns a 64 bit integer (see
http://www.mysql.com/doc/m/y/mysql_insert_id.html)? Though, ignoring
this, you should normally get a steady 0 returned.
Another possible cause of failure could be that you did not recompile
your application and therefore have a version
Hi.
On Mon, Jul 23, 2001 at 04:32:54PM +0400, [EMAIL PROTECTED] wrote:
> Problem: INSERT fields AUTO_INCREMENT passes correctly, and call
> after that mysql_insert_id () all the same returns 0. Linux 2.2.14
> operating system. Mysql version - 2.23.38
Hm. You know that the return value is of type
