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  • 1.sql统计语句

    select exam_item_code, exam_item, EXAMDATE, count(distinct patient_id)
    from (select t2.exam_item_code,
    t2.exam_item,
    to_char(t1.exam_date_time, 'yyyy-mm') as EXAMDATE,
    t1.patient_id
    from exam.exam_master t1, exam.exam_items t2
    where t1.exam_no = t2.exam_no
    and exam_date_time >=
    to_date('2014-07-01 00:00:00', 'yyyy-mm-dd hh24:mi:ss')
    and exam_date_time <=
    to_date('2015-06-30 23:59:59', 'yyyy-mm-dd hh24:mi:ss')
    and exam_class = 'CT'
    and performed_by = '2401'
    and t2.exam_item_code in
    ('210300110', '210300111', '210300109', '210300108',
    '210300036', '210300037', '210300134', '210300125',
    '210300069', '210300070', '210300071', '210300080',
    '210300115')
    --and t2.exam_item_code in ('210300134')
    )
    group by exam_item_code, exam_item, EXAMDATE
    order by exam_item_code, exam_item, EXAMDATE desc

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  • 原文地址:https://www.cnblogs.com/yinsheng/p/5879239.html
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