We all know how to sum up by year select to_char(trunc(hiredate,’Y’),’YYYY’) year, count(*) from scott.emp group by trunc(hiredate,’Y’) order by trunc(hiredate,’Y’) YEAR COUNT(*) —- ——– 1980 1 1981 10 1982 1 1987 2 This is quite easy. Same for day (DD), century (CC), quartal (Q), hour (HH24), month (MM) But how do you group by, […]
TIME_BUCKET group by time period
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