Viewing 15 posts - 121 through 135 (of 196 total)
Hello Denise,
I don't think this is quite difficult. Unfortunately it's more difficult to really understand your example.
Please read the following link (I did too at the beginning 😉 ) and...
June 10, 2013 at 1:30 pm
You should give some more details about your replication.
Publication: one more more?
Articles: one or more? Tables only?
Subscriptions: one ore more? Push or Pull?
Subscriptions create with or without initialization?
Distribution-Agent(s) running?
June 10, 2013 at 1:24 pm
When creating the snapshot the publisher places some locks to know where to start transferring transactions to the distribution database. Then the data for the snapthot has to be read.
When...
June 7, 2013 at 8:48 am
I'm not sure about what you explained.
I think you changed the table on the publisher and want the same changes on the subscriber? If so:
There is an option on the...
June 7, 2013 at 8:40 am
hi,
ending a conversation on the target DOES send a message back to the initiator.
For testing I recommend creating the queue on the initiator without any stored procedure. This way the...
June 7, 2013 at 8:35 am
hi,
there's a quite good select here:
Have a look at it and pick out the data you need 🙂
June 7, 2013 at 8:30 am
hi,
I can't give you a full answer but maybe some hints.
In general you can use the same partitioning schema and function for all tables. The question is what happens if...
June 7, 2013 at 8:28 am
The identity column should not be the problem. The table on the subscriber is created using the usual create-table-statement.
As arnipetursson already mentioned: check the snapshot agent for error messages.
June 7, 2013 at 8:23 am
hi,
if I undestand right your subscriber is fully up to date.
Do you have another subscriber? If you reinitialized the replication maybe you did it for all subscribers and not...
June 7, 2013 at 8:22 am
I'm not sure if there is a function in MySQL.
In T-SQL one way would be to create a calender-table which contains one record for each day for the past and...
May 30, 2013 at 6:05 am
hi,
in SQL you can reference to each object using a 4-part name: [Server].[Database].[Schema].[Table]
[Server] has to be registered as a Linked Server, for details have a look at books online (
May 30, 2013 at 6:03 am
Thank's for your reply.
It looks much more compact than the way I did. But it won't be much faster as the where-conditition has still to be executed twice on the...
April 2, 2013 at 2:50 am
hi Grasshopper,
there is no standard script for such cases. Yes, actually I compared any single table on each subscriber (thank goodness there are only 2).
In sys.objects I can see when...
February 27, 2013 at 12:22 pm
Thank you, exactly what I was looking for 🙂
February 8, 2013 at 8:22 am
hello Tony,
I agree with Grant Fritchey. Maybe I can give you a summarized aspect:
Let's say you have data that you can group by month.
Partitioning is great if you want...
January 25, 2013 at 3:53 pm
Viewing 15 posts - 121 through 135 (of 196 total)