remove filtered data from cube

Question

remove filtered data from cube

Tobias Boehler

Old Hand

Points: 375
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October 12, 2005 at 10:14 am

#166253

hello
is it possible to remove data from a cube?
the background:
my cube should contain actual data
i want to refresh the data for the last 90 days (because of to long loading time)
refreshing by filtering data is not posible
can i now remove data (last 90 days) from the cube and execute an incremental update for 90 days?
or is there any other way to solve this problem
thanks, tobi

Viewing 8 posts - 1 through 7 (of 7 total)

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Bad dog Old Hand Points: 387 More actions · Answer 1

Hi Tobi

Not sure about the filtering / incremental updates but do you know about partitions? If you are running the enterprise version you can partition the cube. Use a new partition for every month or quarter and you can process just one partittion. (This won't solve your immediate problem but could help in the same scenario in future)

Hope this helps

Tobias Boehler Old Hand Points: 375 More actions · Answer 2

Hello Bad dog.

Hello everybody else 😉

One short question to these pratitions. At the moment i try to read everything about partitions, but one thing i can't find: querying the cube.

How does the query work?

Example. I make a partition on [Date].][2005].[Q1]

I now start a query on the cube with Date for example [2005-02-01]:[2005-03-01]

-> Then the query goes to the partition?

Other example: Date [2004-12-01]:[2005-02-01]

-> Then it queries the normal cube?

I think i have to less literature here 🙁

But i need this information urgently...

Thanks in advance, Tobi

stevefromOZ SSC-Forever Points: 43646 More actions · Answer 3

You're on the right track re querying the cube, except that the server decides/knows which partition or partitions to use to resolve a query. Having partitions also allows for multiple processes to access dat in parallel, which can make queries return faster as a result. It is critical however that you set the data slice on the partition to inform the engine of the contents. Note that the slice is *not* the filter for the partition, which needs to be set also.

Steve.

Tobias Boehler Old Hand Points: 375 More actions · Answer 4

Hello again.

Thank you for this informations. Now i have another problem. I made a partition and set the

"Data Slice" on [Date].[2005].[Q1] and the

"Filter Statement" with

CONVERT(DATETIME, '2005-01-01 00:00:00', 102) < = IBIrep_LEP_TimeDimensionTable.datum AND IBIrep_LEP_TimeDimensionTable.datum <= CONVERT(DATETIME, '2005-03-31 23:59:59', 102)

Now i put a query on [Date].[2005].[Q1].[January].[3] and then the AS returns with exactly the doubled values of the real data...

What did i wrong?

I did no incrementel update. I just designed the storage and processed the cube one time.

Thanks again... Tobi

stevefromOZ SSC-Forever Points: 43646 More actions · Answer 5

Hey Tobi,

How many partitions do you have in total now on the cube? If you didn't i) remove the original partition or ii) set it's filter (and slice) [and re-process it] then it is likely that it still contains data for the same period that you created the new partition for.

When doing this testing, you might want to make a copy of the cube (or even the whole catalog) to work with.

Steve.

Tobias Boehler Old Hand Points: 375 More actions · Answer 6

Good morning Steve,

now it dawns on me 🙂

What a great monday.

My _wrong_ thought was:

The cube ignores the other partitions. But of the wrong slicer-statement in the original partition the cube queryed the data from it too and added it to the other values.

Now I'll delete the original partition and create partitions for:

2001 - 2004_____1 year = 1 partition_____(4 partitions)

2005___________1 quater = 1 partition___(4 partitions)

During the day I'll check the data and post my results.

Thank you for this infromation.

Best wishes, Tobi

Tobias Boehler Old Hand Points: 375 More actions · Answer 7

Hello

Short statement to the testings.

I did a lot queries yesterday and had no differences to the origin data.

Thanks for help everybody.

Tobi