SQL中distinct与order by

本文探讨了SQL查询中使用distinct关键字的正确方式,并通过具体例子展示了如何避免orderby与distinct之间的逻辑矛盾,确保查询结果准确无误。

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错误

select distinct ga.id 

from GeneralGranted pro, GeneralMember gm
left join gm.department d left join gm.institute i left join i.type sys left join gm.university u left join pro.application ga left join pro.generalVariation var left join pro.generalVariation var2 left join pro.subtype so

where gm.application.id = ga.id and gm.isDirector=1

group by pro.id, pro.name, gm.member.id, gm.memberName, gm.university.id, gm.agencyName, so.name, ga.disciplineType, ga.year, ga.id, var.id

having COUNT(var2.id) > 0

order by pro.name asc, pro.id asc

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去掉distinct正确

select ga.id 

from GeneralGranted pro, GeneralMember gm
left join gm.department d left join gm.institute i left join i.type sys left join gm.university u left join pro.application ga left join pro.generalVariation var left join pro.generalVariation var2 left join pro.subtype so

where gm.application.id = ga.id and gm.isDirector=1

group by pro.id, pro.name, gm.member.id, gm.memberName, gm.university.id, gm.agencyName, so.name, ga.disciplineType, ga.year, ga.id, var.id

having COUNT(var2.id) > 0

order by pro.name asc, pro.id asc


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加上select pro.name , pro.id正确

select distinct ga.id, pro.name , pro.id

from GeneralGranted pro, GeneralMember gm
left join gm.department d left join gm.institute i left join i.type sys left join gm.university u left join pro.application ga left join pro.generalVariation var left join pro.generalVariation var2 left join pro.subtype so

where gm.application.id = ga.id and gm.isDirector=1

group by pro.id, pro.name, gm.member.id, gm.memberName, gm.university.id, gm.agencyName, so.name, ga.disciplineType, ga.year, ga.id, var.id

having COUNT(var2.id) > 0

order by pro.name asc, pro.id asc


错误原因在于order by与distinct的逻辑矛盾,distinct ga.id后order by pro.name , pro.id错误
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