Best way to check the data

I have a made some changes to an existing view. The change is adding a join and a new column from new table. I have created a version2 with all these changes. I want to make sure that same data exists in both the views. what is the best way to test the data.

Bad wording, the query results could be:
No rows selected. Same number of rows on both views and same data (considering view1 structure and view without the new column)
At least one row selected. If the count(*) on both was different, rows returned will be at least the difference. If the count was the same, then every returned row means that there's no equivalent row on view2 and a data difference may exist in at least one of the fields, so You have to find the equivalent row on view2 to compare.
On the query sintaxis, as view2 hava a new column and I guess it's not a constant value, You have to specify every column for the view1, and the same columns for view2, so new field isn't included and compared.

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www.paragon-software.com
Some general Apple Help www.apple.com/support/
Also,
Mac OS X Help
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Isolating Issues in Mac OS
http://support.apple.com/kb/TS1388
https://www.apple.com/support/osx/
https://www.apple.com/support/quickassist/
http://www.apple.com/support/mac101/help/
http://www.apple.com/support/mac101/tour/
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http://docs.info.apple.com/article.html?artnum=304725
Apple Mac App Store
https://discussions.apple.com/community/mac_app_store/using_mac_apple_store
How to Buy Mac OS X Mountain Lion/Lion
http://www.apple.com/osx/how-to-upgrade/
TimeMachine 101
https://support.apple.com/kb/HT1427
http://www.apple.com/support/timemachine
Mac OS X Community
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Hi,
I am Googling on this and find many different answers and I am not sure what the best approach is. I have two lists:
List A: with 10000 entries (String with length 6 all of them)
List B: with 10002 entries (String with length 6 all of them)
List A and B are almost identical but with 2 differences. I want to find these two differences or in other words generating a third list, List C with these 2 differences.
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Maria1990 wrote:
I think I found the easiest approach...
listA = [a, b, c];
listB = [a, b];
listC = listA.removeAll(listB);
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3. If you find a match, go back to 1.
4. If you exhaust one list, then any remaining elements in the other are also "mismatches".
HIH
Winston
PS: With this approach, you can also flag your mismatches as "A not in B" or "B not in A".
Edited by: YoungWinston on Jul 27, 2010 4:24 AM

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I would recommend AES256.
I would also recommend a VTI tunnel vs. GRE/IPSec. However, both, depending on your IOS, should support AES256.
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