1、在面試的時候碰到一個 問題，就是讓寫一張表中有id和name 兩個字段，查詢出name重復的所有數據，現在列下：

select * from xi a where (a.username) in  (select username from xi group by username  having count(*) > 1)

2、查詢出所有數據進行分組之后，和重復數據的重復次數的查詢數據，先列下：

select  count(username) as '重復次數',username from xi group by username  having count(*)>1 order by username desc

3、一下為 查看別人的 結果，現列下：查詢及刪除重復記錄的方法大全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 peoplewhere 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 awhere (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 awhere (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 awhere (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 integerdeclare cur_rows cursor local for select 主字段,count(*) from 表名 group by 主字段 having count(*) >； 1open cur_rowsfetch cur_rows into @id,@maxwhile @@fetch_status=0beginselect @max = @max -1set rowcount @maxdelete from 表名 where 主字段 = @idfetch cur_rows into @id,@maxendclose cur_rowsset rowcount 0方法二＂重復記錄＂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?！　?、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?！　∪绻摫硇枰獎h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 tableNamedrop table tableNameselect * into tableName from #Tmpdrop 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?！　?、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 tableNameselect min(autoID) as autoID into #Tmp2 from #Tmp group by Name,autoIDselect * 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 tablenamegroup by idhaving count(id) > 1)