SQL SEQUENTIAL NUMBER WITH GROUP , THE OPPOSITE OF PARTITION BY

Question

Hi,
I need to create a Progressive Number on different group in a Select on SQL, the number should be progressive with different group , but the same inside the same group.
Here is the code i use now with the result :

select   
ROW_NUMBER()OVER(PARTITION BY  SD.SALEDOCID ORDER BY sd.DOCNO ASC) AS ProgressivoDocumento,  
sd.SaleDocId ,  
cs.custsupp ,  
'CINV' AS TipoDocument,  
sd.DocNo As NumeroDocumento,  
sd.DocumentDate as DataEmissioneDocumento  
from MA_SaleDoc SD   
JOIN MA_PyblsRcvbls PY on sd.DocNo = py.DocNo  
right join MA_CustSupp CS on cs.CustSupp = sd.CustSupp  
join MA_SaleDocSummary sds on sd.SaleDocId = sds.SaleDocId   
join MA_SaleDocPymtSched sdp on sdp.SaleDocId = sd.SaleDocId  
  
where   
( sd.DocumentType = '3407874'or sd.DocumentType = '3407875') and py.Settled = 0  and sd.DocumentDate >='20210101'  and py.DocumentDate >='20210101' AND CS.CustSuppType = '3211264'  

And this is what i find

But i need exactly the opposite like this :

Can someone help me on how to solve this?
Thank you very much

Answer 1

Try 'dense_rank() over (order by SaleDocId)'.

Answer 2

Hi @daniele franzini ，

Welcome to the microsoft TSQL Q&A forum!

The DENSE_RANK() function mentioned by Viorel-1 can solve your problem. Please refer to his answer. For more details, please refer to:
DENSE_RANK (Transact-SQL)

Regards
Echo