SQL反復記載查詢的幾種辦法。本站提示廣大學習愛好者：（SQL反復記載查詢的幾種辦法）文章只能為提供參考，不一定能成為您想要的結果。以下是SQL反復記載查詢的幾種辦法正文
1、查找表中過剩的反復記載，反復記載是依據單個字段（peopleId）來斷定
select * from people
where peopleId in (select peopleId from people group by peopleId having count
(peopleId) > 1)
2、刪除表中過剩的反復記載，反復記載是依據單個字段（peopleId）來斷定，只留有rowid最小的記載
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、查找表中過剩的反復記載（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、刪除表中過剩的反復記載（多個字段），只留有rowid最小的記載
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、查找表中過剩的反復記載（多個字段），不包括rowid最小的記載
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)
比喻說
在A表中存在一個字段“name”，
並且分歧記載之間的“name”值有能夠會雷同，
如今就是須要查詢出在該表中的各記載之間，“name”值存在反復的項；
Select Name,Count(*) From A Group By Name Having Count(*) > 1
假如還查性別也雷同年夜則以下:
Select Name,sex,Count(*) From A Group By Name,sex Having Count(*) > 1
(三)
辦法一
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
辦法二
有兩個意義上的反復記載，一是完整反復的記載，也即一切字段均反復的記載，二是部門症結字段重
復的記載，好比Name字段反復，而其他字段紛歧定反復或都反復可以疏忽。
1、關於第一種反復，比擬輕易處理，應用
select distinct * from tableName
便可以獲得無反復記載的成果集。
假如該表須要刪除反復的記載（反復記載保存1條），可以按以下辦法刪除
select distinct * into #Tmp from tableName
drop table tableName
select * into tableName from #Tmp
drop table #Tmp
產生這類反復的緣由是表設計不周發生的，增長獨一索引列便可處理。
2、這類反復成績平日請求保存反復記載中的第一筆記錄，操作辦法以下
假定有反復的字段為Name,Address，請求獲得這兩個字段獨一的成果集
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)
最初一個select即獲得了Name，Address不反復的成果集（但多了一個autoID字段，現實寫時可以寫
在select子句中省去此列）
(四)查詢反復
select * from tablename where id in (
select id from tablename
group by id
having count(id) > 1
)
