Thu Apr 21 2022 04:04:25 GMT+0000 (Coordinated Universal Time)
Saved by @Bambibo9799
select * from( select distinct customer_id, dense_rank()over( order by customer_id asc) rn_low, dense_rank()over( order by customer_id desc) rn_high from subscriptions as s ) t1 where rn_low <= 5 or rn_high <= 5 order by customer_id
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