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how can we perform incremental aggregation?explain with
example?
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Answer Posted / deepak
I wish people answering in this forum use correct English.
Many sentences are confusing at first glance due to poor
grammer.
| Is This Answer Correct ? | 8 Yes | 20 No |
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rank() over (partition by opt2.dim_plat_site_id, opt2.dim_site_opt_sid order by case when opt2.dm_market_flg in ('Y', 'U') then 1 else 2 end, lkp.contact_rank) as rank1, case opt2.contact_type when 'Buyer' then row_number() over (partition by opt2.dim_plat_site_id, opt2.dim_site_opt_sid, lkp.contact_rank order by has_name_flg desc, ship_to_flg desc , last_order_dt desc) when 'Decision Maker' then row_number() over (partition by opt2.dim_plat_site_id, opt2.dim_site_opt_sid, lkp.contact_rank order by has_name_flg desc , last_quote_dt desc , mailability_score desc , source_ranking desc) when 'Influencer' then row_number() over (partition by opt2.dim_plat_site_id, opt2.dim_site_opt_sid, lkp.contact_rank order by has_name_flg desc, mailability_score desc, source_ranking desc) when 'Payer' then row_number() over (partition by opt2.dim_plat_site_id, opt2.dim_site_opt_sid, lkp.contact_rank order by has_name_flg desc, mailability_score desc, source_ranking desc) --elu 05/28/2013 else row_number() over (partition by opt2.dim_plat_site_id, opt2.dim_site_opt_sid, lkp.contact_rank order by has_name_flg desc, mailability_score desc, source_ranking desc) end rank2 row_number() over (partition by opt3.dim_plat_site_id, opt3.dim_site_opt_sid order by rank1,rank2) as "rank", case when "rank"<= opt3.maximum_value then 'Y' else 'N' end as include_flg
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