1 、查找表中多余的重復記錄，重復記錄是根據單個字段（peopleId）來判斷
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記錄，重復記錄是根據單個字段（peopleId）來判斷，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記錄（多個字段）?
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記錄（多個字段），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記錄（多個字段），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
在A表中存在一個字段“name”，
而且不同記錄之間的“name”值有可能會相同，
現在就是需要查詢出在該表中的各記錄之間，“name”值存在重復的項；
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

有兩個意義上的重復記錄，一是完全重復的記錄，也即所有字段均重復的記錄，

二是部分關鍵字段重復的記錄，比如Name字段重復，而其他字段不一定重復或都重復可以忽略。

1 、對于第一種重復，比較容易解決，使用
select ? distinct ? * ? from ?tableName

　　就可以得到無重復記錄的結果集。

　　如果該表需要刪除重復的記錄（重復記錄保留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

　　發生這種重復的原因是表設計不周產生的，增加唯一索引列即可解決。

2 、這類重復問題通常要求保留重復記錄中的第一條記錄，操作方法如下

　　假設有重復的字段為Name,Address，要求得到這兩個字段唯一的結果集
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select即得到了Name，Address不重復的結果集（但多了一個autoID字段，實際寫時可以寫

在select子句中省去此列）

(四)查詢重復
select ? * ? from ?tablename

where ?id? in ?( select ?id? from ?tablename? group ? by ?id? having ? count (id)? > ? 1 )