oracle 根据时间粒度统计记录数

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最新推荐文章于 2024-09-18 17:59:46 发布

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oracle 根据时间粒度统计记录数

select count(a.Primary_key_column) as receive, b.period
    from ((select Primary_key_column from tableName) a
          right join
          (select level, trunc(sysdate, 'hh') - 1 / 24 * level as period
             from dual
            connect by level <= 48
            order by level) b on trunc(a.Primary_key_column, 'hh') = b.period)
   group by period
   order by period desc