Reset Next Range Starting Value?

Hello,
Is there any way to reset the "Next range starting value" on a table
with Identity Range management?
For some reason, the next range starting value has gone too high. The
maximum identity is 32767, and the next range starting value is 32650.
I did have the range size at subscribers set to 200, but there wasn't
enough room to give that range so the initial replication failed. I
changed it to 50 in order to get the initial subscription pulling.
The actual maximum identity in the table is less than 2000. I only
have 2 subscribers.
I'm not sure how it got so high, but is there a way to reset it?
Thanks,
Jeff
You can manually update the msrepl_identity (sp?) table in the distribution
database.
However, you should poll your subscribers and publisher to find out what is
in effect there.
Check the check constraint on the identity key and do a DBCC
checkindent('tablename') to get the current value and the value in use.
Hilary Cotter
Looking for a SQL Server replication book?
http://www.nwsu.com/0974973602.html
Looking for a FAQ on Indexing Services/SQL FTS
http://www.indexserverfaq.com
"Jeff Hedlund" <jhedlund@.gmail.com> wrote in message
news:1116082608.807565.267540@.o13g2000cwo.googlegr oups.com...
> Hello,
> Is there any way to reset the "Next range starting value" on a table
> with Identity Range management?
> For some reason, the next range starting value has gone too high. The
> maximum identity is 32767, and the next range starting value is 32650.
> I did have the range size at subscribers set to 200, but there wasn't
> enough room to give that range so the initial replication failed. I
> changed it to 50 in order to get the initial subscription pulling.
> The actual maximum identity in the table is less than 2000. I only
> have 2 subscribers.
> I'm not sure how it got so high, but is there a way to reset it?
> Thanks,
> Jeff
>

repost on subtracting date range data between tables

Hi everyone. I posted this previously but made an error in my original
posting. So i am posting again with corrections in place, in hopes someone
might shed light on a possible solution.
What i have is two tables containing date range date. I need to find the
difference between this date. So for example if i have:
table a:
[start] [finish]
1 10
18 19
23 26
28 31
table b:
[start] [finish]
4 5
18 18
25 28
Then the result of table a - table b:
[start] [finish]
1 3
6 10
19 19
23 24
29 31
in effect, if a date range in table_b intersects a date range in table_a,
then that data in the intersection is removed from the date range in table_a
for example, the following are example cases where we subtract from a range
in table_a where there are ranges in table_b which intersect with that range
in table_a:
[From Table A ] - [ From Table B]:
1. {23....37} - {25...29} = {23...24}
2. {23....37} - {26...32} = {23...25}, {33...37}
3. {23....37} - [ {25...27} , {31...33} ] = {23...24}, {28...30},
{34...37}
I am considering solving this problem by using: A - (A AND B). Here A AND B
is a pure subset of A. Maybe this would make the problem easier to solve.
Any help on this would be really appreciated!
Many thanks.
peter-- There are probably lots of ways of doing this, the code
-- below uses recursive CTEs (you'll need SQL Server 2005)
CREATE TABLE TabA(start INT, finish INT)
INSERT INTO TabA(start,finish)
SELECT 1, 10 UNION ALL
SELECT 18, 19 UNION ALL
SELECT 23, 26 UNION ALL
SELECT 28, 31;
CREATE TABLE TabB(start INT, finish INT)
INSERT INTO TabB(start, finish)
SELECT 4, 5 UNION ALL
SELECT 18, 18 UNION ALL
SELECT 25, 28;
WITH CTE_Recur(startgrp,finishgrp,level,start
,finish) AS
(
SELECT a.start,a.finish,1,a.start,a.finish
FROM TabA a
WHERE EXISTS (SELECT * FROM TabB b WHERE a.start<=b.finish AND
a.finish>=a.start)
AND NOT EXISTS (SELECT * FROM TabB b WHERE a.start>=b.start AND
a.finish<=b.finish)
UNION ALL
SELECT a.startgrp,a.finishgrp,level+1,a.start,b.start-1
FROM CTE_Recur a
INNER JOIN TabB b ON a.start<b.start AND a.finish>=b.start
UNION ALL
SELECT a.startgrp,a.finishgrp,level+1,b.finish+1,a.finish
FROM CTE_Recur a
INNER JOIN TabB b ON a.finish>b.finish AND a.start<=b.finish
),
CTE_Leaves(startgrp,finishgrp,level) AS
(
SELECT startgrp,finishgrp,max(level)
FROM CTE_Recur
GROUP BY startgrp,finishgrp
)
SELECT r.start,r.finish
FROM CTE_Recur r
INNER JOIN CTE_Leaves l ON l.startgrp=r.startgrp AND
l.finishgrp=r.finishgrp AND l.level=r.level
UNION
SELECT a.start,a.finish
FROM TabA a
WHERE NOT EXISTS (SELECT * FROM TabB b WHERE a.start<=b.finish AND
a.finish>=a.start)
ORDER BY 1,2|||hi, thanks for that.but would you know how to write this is sql 2000?
much appreciated
peter
<markc600@.hotmail.com> wrote in message
news:1137272625.701449.190200@.g43g2000cwa.googlegroups.com...
> -- There are probably lots of ways of doing this, the code
> -- below uses recursive CTEs (you'll need SQL Server 2005)
>
> CREATE TABLE TabA(start INT, finish INT)
> INSERT INTO TabA(start,finish)
> SELECT 1, 10 UNION ALL
> SELECT 18, 19 UNION ALL
> SELECT 23, 26 UNION ALL
> SELECT 28, 31;
> CREATE TABLE TabB(start INT, finish INT)
> INSERT INTO TabB(start, finish)
> SELECT 4, 5 UNION ALL
> SELECT 18, 18 UNION ALL
> SELECT 25, 28;
> WITH CTE_Recur(startgrp,finishgrp,level,start
,finish) AS
> (
> SELECT a.start,a.finish,1,a.start,a.finish
> FROM TabA a
> WHERE EXISTS (SELECT * FROM TabB b WHERE a.start<=b.finish AND
> a.finish>=a.start)
> AND NOT EXISTS (SELECT * FROM TabB b WHERE a.start>=b.start AND
> a.finish<=b.finish)
> UNION ALL
> SELECT a.startgrp,a.finishgrp,level+1,a.start,b.start-1
> FROM CTE_Recur a
> INNER JOIN TabB b ON a.start<b.start AND a.finish>=b.start
> UNION ALL
> SELECT a.startgrp,a.finishgrp,level+1,b.finish+1,a.finish
> FROM CTE_Recur a
> INNER JOIN TabB b ON a.finish>b.finish AND a.start<=b.finish
> ),
> CTE_Leaves(startgrp,finishgrp,level) AS
> (
> SELECT startgrp,finishgrp,max(level)
> FROM CTE_Recur
> GROUP BY startgrp,finishgrp
> )
> SELECT r.start,r.finish
> FROM CTE_Recur r
> INNER JOIN CTE_Leaves l ON l.startgrp=r.startgrp AND
> l.finishgrp=r.finishgrp AND l.level=r.level
> UNION
> SELECT a.start,a.finish
> FROM TabA a
> WHERE NOT EXISTS (SELECT * FROM TabB b WHERE a.start<=b.finish AND
> a.finish>=a.start)
> ORDER BY 1,2
>|||-- This isn't quite equivalent and has some restrictions
-- such as table B must not have any overlapping ranges,
-- but will work on SQL Server 2000
CREATE TABLE TabA(start INT, finish INT)
INSERT INTO TabA(start,finish)
SELECT 1, 10 UNION ALL
SELECT 18, 19 UNION ALL
SELECT 23, 26 UNION ALL
SELECT 28, 31;
CREATE TABLE TabB(start INT, finish INT)
INSERT INTO TabB(start, finish)
SELECT 4, 5 UNION ALL
SELECT 18, 18 UNION ALL
SELECT 25, 28;
SELECT COALESCE((SELECT MAX(b2.finish)+1 FROM TabB b2 WHERE b2.finish <
b.start and b2.finish > a.start),a.start) as start,
b.start-1 as finish
FROM TabA a
INNER JOIN TabB b ON a.start<b.start AND a.finish>=b.start
WHERE EXISTS (SELECT * FROM TabB b WHERE a.start<=b.finish AND
a.finish>=a.start)
AND NOT EXISTS (SELECT * FROM TabB b WHERE a.start>=b.start AND
a.finish<=b.finish)
UNION
SELECT b.finish+1,
COALESCE((SELECT MIN(b2.start)-1 FROM TabB b2 WHERE b2.start >
b.finish and b2.start < a.finish),a.finish)
FROM TabA a
INNER JOIN TabB b ON a.finish>b.finish AND a.start<=b.finish
WHERE EXISTS (SELECT * FROM TabB b WHERE a.start<=b.finish AND
a.finish>=a.start)
AND NOT EXISTS (SELECT * FROM TabB b WHERE a.start>=b.start AND
a.finish<=b.finish)
UNION
SELECT a.start,a.finish
FROM TabA a
WHERE NOT EXISTS (SELECT * FROM TabB b WHERE a.start<=b.finish AND
a.finish>=a.start)
ORDER BY 1,2|||On Sun, 15 Jan 2006 04:45:27 +1000, peter walker wrote:

>Hi everyone. I posted this previously but made an error in my original
>posting. So i am posting again with corrections in place, in hopes someone
>might shed light on a possible solution.
(snip)
Hi Peter,
I just posted a reply to your first message about this problem.
Hugo Kornelis, SQL Server MVP|||yet another approach would be to use a calendar table:
select [date] date_in_range from calendar c
where exists(select 1 from a
where c.[date] between a.[start] and a.[finish])
and not exists(select 1 from b
where c.[date] between b.[start] and b.[finish])
The query will return a set of dates. If you need intervals, that's
also quite easy to accomplish