Para-dox wrote:
>
> I'm using 3.22.25
>
> ----- Original Message -----
> From: Christian Mack <Mack@stripped>
> To: Para-dox <paradox@stripped>
> Cc: <mysql@stripped>
> Sent: Tuesday, July 27, 1999 2:33 PM
> Subject: Re: Subtraction vapor
>
> > Para-dox wrote:
> > >
> > > Something seems to be causing inconsistancy...
> > >
> > > SELECT (Sum(PAYMENTS.AMT)/100) FROM INVOICES LEFT JOIN PAYMENTS ON
> > > PAYMENTS.INVNUM = INVOICES.INVNUM WHERE INVOICES.INVNUM = 133674 GROUP
> BY
> > > INVOICES.INVNUM;
> > >
> > > This returns one row with 134.56
> > >
> > > SELECT (((((10000+INVOICES.TAXPERC+IF(INVOICES.HNDTYPE=0,
> INVOICES.HNDVAL,
> > > 0))*INVOICES.SUBTOTAL/10000)+IF(INVOICES.HNDTYPE<>0,
> INVOICES.HNDVAL,
> > > 0)))/100) FROM INVOICES LEFT JOIN PAYMENTS ON PAYMENTS.INVNUM =
> > > INVOICES.INVNUM WHERE INVOICES.INVNUM = 133674 GROUP BY INVOICES.INVNUM;
> > >
> > > This returns one row with 134.56
> > >
> > > SELECT (((((10000+INVOICES.TAXPERC+IF(INVOICES.HNDTYPE=0,
> INVOICES.HNDVAL,
> > > 0))*INVOICES.SUBTOTAL/10000)+IF(INVOICES.HNDTYPE<>0,
> INVOICES.HNDVAL,
> > > 0)))/100) - (Sum(PAYMENTS.AMT)/100) FROM INVOICES LEFT JOIN PAYMENTS ON
> > > PAYMENTS.INVNUM = INVOICES.INVNUM WHERE INVOICES.INVNUM = 133674 GROUP
> BY
> > > INVOICES.INVNUM;
> > >
> > > This *should* return 0, I believe, but instead it returns -134.56 ??
> > > Am I forgetting something?
> > >
> > > Regards, Dave (paradox@stripped)
> >
> > Hi Dave
> >
> > The syntax looks OK for me.
> > I also think it should return 0.
> > What version do you use?
> > I heard on this list about a GROUP BY problem in 3.23.
> > If you have this one, try to apply the patch from the mailinglist archive.
> >
> > Tschau
> > Christian
> >
> >
Hi Dave
Seems there is an GROUP BY error in 3.22.25 too.
Tschau
Christian
