Re: RECORD COMPARISON
- From: "Bmack500" <brett.mack@xxxxxxxxx>
- Date: 14 Feb 2007 05:57:43 -0800
The changes are not committed; currently I'm just clearing the
database and writing fresh. It's a database I use to collect server
information through WMI.
I would just like to implement a change notification mechanism; i.e.,
a new local users is created, deleted; the duplex settings on a nic
are changed; etc....
So I just need to compare the existing data with the new incoming
data, and decide what to do from there. There will be no users making
constant changes.
On Feb 13, 1:34 pm, "RobinS" <Rob...@xxxxxxxxxxxxxxx> wrote:
Well, here is where I was going. If you have a dataset and the users have
made changes to it, and you have *not* committed the changes, you can
easily figure out what rows were changed, and what the changes were. So are
you saying that you have *not* committed the (new) changes yet?
Is your data in DataTables or DataSets (not that it really matters, I'm
just asking).
Robin S.
Ts'i mahnu uterna ot twan ot geifur hingts uto.
-----------------------------------------------"Bmack500" <brett.m...@xxxxxxxxx> wrote in message
news:1171372467.607538.275600@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
The original data is already saved to the database, but I could save
the new data to a second db or whatever it takes to do a valid
comparison.
It's sql server 2000 enterprise, not 2005 (unfortunately).
On Feb 12, 9:49 pm, "RobinS" <Rob...@xxxxxxxxxxxxxxx> wrote:
Have you already saved the changes to the database?
Robin S.
----------------------------------"Bmack500" <brett.m...@xxxxxxxxx>
wrote in message
news:1171315998.061772.157400@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
I need to rapidly compare an old record with it's newest version, in
order to general alerts. My code is in vb.net (.net 2.0).
The record has many related tables; what is going to be the quickest
way to compare the records? Almost every field will need to be
compared, and also related records; deletion and additions.
I'd just hate to do it by iterating every field of every record....
Thanks in advance!
.
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