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  • oracle去重操作

    上次用过忘记了,所以备份一下

    在几千条记录里,存在着些相同的记录,如何能用SQL语句,删除掉重复的呢
    1、查找表中多余的重复记录,重复记录是根据单个字段(peopleId)来判断
    select * from people
    where peopleId in (select peopleId from people group by peopleId having count(peopleId) > 1)

    2、删除表中多余的重复记录,重复记录是根据单个字段(peopleId)来判断,只留有rowid最小的记录
    delete from people
    where peopleName in (select peopleName from people group by peopleName having count(peopleName) > 1)
    and peopleId not in (select min(peopleId) from people group by peopleName having count(peopleName)>1)

    3、查找表中多余的重复记录(多个字段)
    select * from vitae a
    where (a.peopleId,a.seq) in (select peopleId,seq from vitae group by peopleId,seq having count(*) > 1)

    4、删除表中多余的重复记录(多个字段),只留有rowid最小的记录
    delete from vitae a
    where (a.peopleId,a.seq) in (select peopleId,seq from vitae group by peopleId,seq having count() > 1)
    and rowid not in (select min(rowid) from vitae group by peopleId,seq having count(
    )>1)

    5、查找表中多余的重复记录(多个字段),不包含rowid最小的记录
    select * from vitae a
    where (a.peopleId,a.seq) in (select peopleId,seq from vitae group by peopleId,seq having count() > 1)
    and rowid not in (select min(rowid) from vitae group by peopleId,seq having count(
    )>1)

    6.消除一个字段的左边的第一位:

    update tableName set [Title]=Right([Title],(len([Title])-1)) where Title like '村%'

    7.消除一个字段的右边的第一位:

    update tableName set [Title]=left([Title],(len([Title])-1)) where Title like '%村'

    8.假删除表中多余的重复记录(多个字段),不包含rowid最小的记录
    update vitae set ispass=-1
    where peopleId in (select peopleId from vitae group by peopleId

    大佬给我的哈哈哈哈 收藏了
    另外还有一个
    https://blog.csdn.net/u012860938/article/details/50791960

    附带我的一个例子
    select count()
    from combine_list
    where list_status = 4
    and to_char(create_date, 'yyyy-mm-dd') not in
    ('2019-10-28', '2019-10-29')
    and (combine_code, component_code) in
    (select combine_code, component_code
    from combine_list t
    where t.list_status = '4'
    group by combine_code, component_code
    having count(
    ) > 1)
    order by combine_code, component_code desc

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  • 原文地址:https://www.cnblogs.com/hbym/p/11758455.html
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