mysql - sql测试条件如果失败转移到下一个人

Question

我正在尝试获取没有 status_mvf = 88 的骨灰盒

status_mvf    urn        desc        date
88            104        no show     2017-03-06
9             104        late        2017-03-07
2             104        on time     2017-03-08
2             210        on time     2017-03-06
2             210        on time     2017-03-07
9             320        late        2017-03-06

SELECT   
    a.Status_Mvf,
    p.URN,
    v.ADMIT_DATE


FROM APPOINTMENT  
WHERE  Status_Mvf <> '88'

--group by p.urn,

ORDER BY URN, STATUS_MVF 

期望的结果 - 没有 104 瓮，因为它有 88

status_mvf   urn       desc      date
2            210       on time   2017-03-06
2            210       on time   2017-03-07
9            320       late      2017-03-06

Answer 1

您可以使用不存在:

select a.*
from appointment a
where not exists (select 1 from appointment a2 where a2.urn = a.urn and a2.status_mvf = '88');