1、查找表中多余的重復(fù)記錄，重復(fù)記錄是根據(jù)單個(gè)字段（peopleId）來(lái)判斷
select * from people
where peopleId in (select?? peopleId from?? people group by?? peopleId having count (peopleId) > 1)
2、刪除表中多余的重復(fù)記錄，重復(fù)記錄是根據(jù)單個(gè)字段（peopleId）來(lái)判斷，只留有rowid最小的記錄
delete from people
where peopleId in (select?? peopleId from people group by?? peopleId?? having count (peopleId) > 1)
and rowid not in (select min(rowid) from?? people group by peopleId having count(peopleId )>1)
3、查找表中多余的重復(fù)記錄（多個(gè)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、刪除表中多余的重復(fù)記錄（多個(gè)字段），只留有rowid最小的記錄
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、查找表中多余的重復(fù)記錄（多個(gè)字段），不包含rowid最小的記錄
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)
(二)
比方說(shuō)
在A表中存在一個(gè)字段“name”，
而且不同記錄之間的“name”值有可能會(huì)相同，
現(xiàn)在就是需要查詢出在該表中的各記錄之間，“name”值存在重復(fù)的項(xiàng)；
Select Name,Count(*) From A Group By Name Having Count(*) > 1
如果還查性別也相同大則如下:
Select Name,sex,Count(*) From A Group By Name,sex Having Count(*) > 1
(三)
方法一
declare @max integer,@id integer
declare cur_rows cursor local for select 主字段,count(*) from 表名 group by 主字段 having
count(*) >； 1
open cur_rows
fetch cur_rows into @id,@max
while @@fetch_status=0
begin
select @max = @max -1
set rowcount @max
delete from 表名 where 主字段 = @id
fetch cur_rows into @id,@max
end
close cur_rows
set rowcount 0
方法二
有兩個(gè)意義上的重復(fù)記錄，一是完全重復(fù)的記錄，也即所有字段均重復(fù)的記錄，
二是部分關(guān)鍵字段重復(fù)的記錄，比如Name字段重復(fù)，而其他字段不一定重復(fù)或都重復(fù)可以忽略。
1、對(duì)于第一種重復(fù)，比較容易解決，使用
select distinct * from tableName
就可以得到無(wú)重復(fù)記錄的結(jié)果集。
如果該表需要?jiǎng)h除重復(fù)的記錄（重復(fù)記錄保留1條），可以按以下方法刪除
select distinct * into #Tmp from tableName
drop table tableName
select * into tableName from #Tmp
drop table #Tmp
發(fā)生這種重復(fù)的原因是表設(shè)計(jì)不周產(chǎn)生的，增加唯一索引列即可解決。
2、這類重復(fù)問(wèn)題通常要求保留重復(fù)記錄中的第一條記錄，操作方法如下
假設(shè)有重復(fù)的字段為Name,Address，要求得到這兩個(gè)字段唯一的結(jié)果集
select identity(int,1,1) as autoID, * into #Tmp from tableName
select min(autoID) as autoID into #Tmp2 from #Tmp group by Name,autoID
select * from #Tmp where autoID in(select autoID from #tmp2)
最后一個(gè)select即得到了Name，Address不重復(fù)的結(jié)果集（但多了一個(gè)autoID字段，實(shí)際寫(xiě)時(shí)可以寫(xiě)
在select子句中省去此列）
(四)查詢重復(fù)
select * from tablename
where id in (select id from tablename group by id having count(id) > 1)
(五)刪除重復(fù)
delete from test a where a.rowid !=
?( select max(b.rowid) from test b where a.id = b.id? )
(六)刪除重復(fù)
方法一：delete from demo a where a.rowid <> (select max(rowid) from demo b where
b.object_id=a.object_id);
耗時(shí)：幾個(gè)小時(shí)以上
方法二： delete from demo where rowid in
(select rid from
(select rowid rid,row_number() over(partition by object_id order by rowid) rn
from demo)
where rn <> 1 );
耗時(shí)：30秒
方法三： create table demo2 as
select object_id,owner... from
(select demo.*,row_number() over(partition by object_id order by rowid) rn from demo)
where rn = 1;
truncate table demo; insert into demo select * from demo2; drop table demo2;
共耗時(shí)： 10秒，適合大數(shù)據(jù)量的情況，產(chǎn)生更少回滾量；