"Any UPDATE or INSERT will always commit the current transaction." ???

If this is correct(which I doubt) then you do not have transactions at all.
As I know
the idea of transaction is to run multiple statements in a way that they are
independent form
the other statements run at the same time?

-- 
eng. Ilian Iliev
Web Software Developer

Mobile: +359 88 66 08 400
Website: http://ilian.i-n-i.org

On Thu, Sep 15, 2011 at 11:43 AM, Tom Evans <tevans...@googlemail.com>wrote:

> On Wed, Sep 14, 2011 at 6:57 PM, Ilian Iliev <il...@i-n-i.org> wrote:
> > Hi Tom,
> > I checked the link you send and probably you are right but
> > can you explain why this is a bad solution?
> > Yes using autocommit() or disabling transactions seems more right
> > are there any downsides of the update() that you are seeing?
> > Image a case when you want to update the result set without breaking the
> > transaction(is this possible)?
> > I am not trolling just asking?
> > Regards,
> > Ilian Iliev
>
> It's bad for future maintenance. Other developers may not know that
> trick, and be confused why you are doing a 'pointless' update, without
> grokking the unannounced side effect. It is always better to be
> explicit rather than implicit.
>
> Any UPDATE or INSERT will always commit the current transaction.
>
> Cheers
>
> Tom
>
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