Hello again,
there are some problems which are present since the migration of firebird
3.0 RC2. Two of them I have reportet, does anyone know where now some
records locked and ibexpert can change them?
thanks
Von: firebird-support@yahoogroups.com
Hi Karol! At the risk of this being confusing or even incorrect (I trust
Dmitry or Ann will correct if it is incorrect).
>SELECT * FROM dbo.XXX X WHERE X.A BETWEEN 2 AND 30 *AND* X.B BETWEEN 5
AND 60
My understanding of this is that Firebird (in theory) have two choices.
Either
(a) use
On Mon, Mar 14, 2016 at 3:41 AM, liviuslivius liviusliv...@poczta.onet.pl
[firebird-support] wrote:
> W dniu 2016-03-14 08:36:40 użytkownik Dmitry Yemanov
> dim...@users.sourceforge.net [firebird-support] <
> firebird-support@yahoogroups.com> napisał:
>
>
> >
On Mon, Mar 14, 2016 at 9:58 AM, Ray Cote
wrote:
> I’ve run into a problem while porting code over from the old
> KInterbasDB library to the new FDB Python library. Receiving the error
> “-817 Metadata update statement not allowed any the current database SQL
>
14.03.2016 10:41, liviuslivius wrote:
> Why?
> Index is a Tree? And if i found VALUE 2 in A key then i can fast find
> value 5 in sub key (leaf)
> You scan throught keys in A, and then in finded nodes you look for leafs
> in B
Compound index key is a single concatenated value, not two separate
Hello:
I’ve run into a problem while porting code over from the old
KInterbasDB library to the new FDB Python library. Receiving the error
“-817 Metadata update statement not allowed any the current database SQL
dialect 3” when attempting to create a table with NUMERIC(15, 3).
My understanding
Hello,
fb3 rc2, newest odbc driver.. If I have to change some records, it works
fine with ibexpert, but with odbc connection (linked tables) I get for he
same records an lock conflict. No problem with the old version 2.5 or 2.1.
Do you know which time firebird 3 final will realeased?
W dniu 2016-03-14 08:36:40 użytkownik Dmitry Yemanov
dim...@users.sourceforge.net [firebird-support]
napisał:
14.03.2016 10:32, liviuslivius wrote:
>
> simple table
> CREATE TABLE XXX(
> A INTEGER
> , B INTEGER
> , CONSTRAINT UK_XXX__A__B UNIQUE(A, B)
> )
14.03.2016 10:32, liviuslivius wrote:
>
> simple table
> CREATE TABLE XXX(
> A INTEGER
> , B INTEGER
> , CONSTRAINT UK_XXX__A__B UNIQUE(A, B)
> )
> commit;
> SELECT * FROM dbo.XXX X WHERE X.A BETWEEN 2 AND 30 *AND* X.B BETWEEN 5
> AND 60
> Select Expression
> -> Filter
> -> Table
Hi,
simple table
CREATE TABLE XXX(
A INTEGER
, B INTEGER
, CONSTRAINT UK_XXX__A__B UNIQUE(A, B)
)
commit;
SELECT * FROM dbo.XXX X WHERE X.A BETWEEN 2 AND 30 AND X.B BETWEEN 5 AND 60
Select Expression
-> Filter
-> Table "XXX" as "X" Access By ID
-> Bitmap
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