sql - Get the row at 60% of the table -
i'd select row @ 60% of table oracle. can find number of 60% row with:
select round(count(*)*0.60) sira (select to_date(time) tarih,lenght hiz table order length desc)
i'm looking name column data row @ 60%. is, 60% of rows should have higher length selected row.
for example, data:
name time length r1 10:00 1 r2 10:02 2 r3 10:04 3 ... r10 10:20 10
i'm looking query prints r4
(row @ 60% ordered decreasing length.)
select * ( select row_number() on (order yt.length desc) rn , count(*) on () cnt , yt.* yourtable yt ) subqueryalias rn = round(cnt * 0.60)
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