架构:
radio_bin.id
radio.id
radio.radio_bin -> radio_bin.id
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查询:
select *
from radio_bin
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72行.
select *
from radio_bin
where id in (select radio_bin from radio)
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50行.
(和FWIW :)
select distinct radio_bin
from radio
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51行,包括null.
这一切都很好.现在:
select *
from radio_bin
where id not in (select radio_bin from radio)
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0行.
为什么?我不应该得到没有收音机指向他们的22个radio_bin.id号码吗?
试试这个,你有一个空值,null不等于任何甚至不是另一个null
select *
from radio_bin
where id not in (select radio_bin from radio where radio_bin is not null)
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另请参阅NOT IN和NULLS,其中显示了如何使用LEFT JOIN或NOT EXISTS
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