How to find tags having duplicate record in the database
When you mentioned duplicate records, you are talking about duplicate record numbers, right?
Identifying points with duplicate record numbers is not very easy, but I'd try first checking the Point Database and filtering by recno attribute
If not possible, the following known issue describes a procedure that could help with your question:
16543OSI8 - -10722 error during point delete can lead to duplicate recno
Furthermore, the snapfix command could help you with duplicate record numbers:
KB00055 - How to repair the snapshot table piarcmem.dat (pibasess -snapfix)
Anyway, I'd suggest you to contact TechSupport in case you find duplicate record numbers on your PI Data Archive so they can help you over a remote session.
Can you elaborate on what you mean by duplicate record?
Are you talking about multiple events with the exact same time, or is it exact time AND exactly the same value?
A quick way to find multiple events with the same timestamp would be to use the PI OLEDB Classic provider (PI SQL Commander). If you query the piarchive..picomp2 table there is a column called "_index", which is > 0 if there are multiple events stored at that point in time.
SELECT * FROM piarchive..picomp2 WHERE time > 't' and _index > 0
would return all events for all tags in the archive, stored at the same time today. Then you should be able to continue building more logic into the query to return what you want in more detail.
Thanks lot for your reply
My question is :-
During a random check I found few tags having duplicate values in the database.
I want to identify those tags having duplicate records in the database.
Please accept my apologies but I still have a very blurry picture of what you understand as duplicate values. On the other hand Asle and / or Luis, may have answered your question already.
If you consider your question unanswered, please identify the pattern that you have in mind when talking about duplicate values. Otherwise, please be so kind and mark the Correct Answer.
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