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How selective deletion works on partition for data targets?

Former Member
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175

Hi BI Guru,

There is new function in NW2004S for the selective deletion on the data targets. It is 'work on partition'on the selective deletion screen. I did the testing on the function, It seems it does not delete the partition, but delete the data from the partition. Does it make sense for your guys? I would expect that drop complete partitions which matches a subset of the records. Could you please tell me how it works? Thanks in advance.

Regards,

Liu Jia

Accepted Solutions (1)

Accepted Solutions (1)

klaus_werner
Active Participant
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Hi Liu Jia,

if the selection of your deletion comprises an entire partition, this partition is dropped. Your expectation is correct. What are your selection criteria for deletion and what is your partitioning criteria?

Regards, Klaus

Former Member
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Hi Klaus,

Thanks for the reply.

My deletion selection criteria is 001.2005. My partitioning criteria is ,

the lower value is 001.2005, the higher value is 004.2005 .

After I deleted the data , I use SE14 to check the partition, the structure is still there. Is the function dropped the partition for 001.2005 then create new one for 001.2005? How can I check the data content for the partition?

Regards,

LiuJIA

klaus_werner
Active Participant
0 Kudos

Hi LiuJIA,

since you chose the lowest value and the partitioning criterion is <= 001.2005 the partition can not be dropped because there may be e.g. 001.2004 in it. That's is probably the reason. Can you test with 002.2005?

Regards, Klaus

klaus_werner
Active Participant
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Hi LiuJIA,

I didn't read thoroughly. In addition to what I've written before: Yes, the structure should be recreated, otherwise your partitioning criteria would have to be changed, so if you see the structure in SE14, there is no problem.

The best way to find out whether the partition is actually dropped is run an SQL trace with ST05 while performing the selective delete.

Regards, Klaus

Former Member
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Hi Klaus,

Thanks for the emails.

I deleted the data for 002.2005 and active SQL trace. I can only understand the comment in the trace list,

-


ALTER TABLE "/BIC/EZPARTI" TRUNCATE PARTITION "/BIC/EZPARTI12005003"

-


it seems drop the partition for 2005003, not 2005002. Does it make sense?

Kind Regards,

Liu Jia

klaus_werner
Active Participant
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Yes it does make sense. That explains why the partition is still there, it is truncated (and not dropped). The partitioning criterion is < ... (I think I've written <= in a previous answer, which was incorrect). So < 2005003 means 2005002, that's correct. Now you could also test what happens when you delete 2005001, then we'd know everything:-)

Regards, Klaus

Former Member
0 Kudos

Hi Klaus,

I did the testing to delete 2005001...

As you expected, I found the following message in ST05.

ALTER TABLE "/BIC/EZPARTI" TRUNCATE PARTITION "/BIC/EZPARTI12005002"

I know the partition behavior now. Thanks. I close the message.

Kind Regards,

Liu Jia

Answers (1)

Answers (1)

Former Member
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Hi Liu Jia,

Could you please let me know where and how can i get the

new function( It is 'work on partition'on the selective deletion screen) you talking about?

we partitioned inventory cube 0ic_c03 by 0calmonth, and range is from 01/2002 to 12/2008. and all the history data have been compressed.

I try to use "slective deletion" in the "contents" tab in cube, and delete the 01/2003 data for example, but in the "delete selections", there is no date infomation (like calmonth, or calday) for me to put, is that the case, how can i delete the 01/2003 data?

Please advice,

Ping

klaus_werner
Active Participant
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I don't know 100 percent sure but I would guess that with an inventory InfoCube you can't delete selectively by time because that would spoil consistency.