Oracle刪除重復行傳智播客
2024-08-29 13:53:37
供稿：網友
查詢及刪除重復記錄的SQL語句
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)
注:rowid為Oracle自帶不用該.....
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
＂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。
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?！ ∪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 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)
=======================================
1。用rowid方法
據據Oracle帶的rowid屬性，進行判定，是否存在重復,語句如下：
查數據:
select * from table1 a where rowid !=(select max(rowid)
from table1 b where a.name1=b.name1 and a.name2=b.name2......)
刪數據：
delete from table1 a where rowid !=(select max(rowid)
from table1 b where a.name1=b.name1 and a.name2=b.name2......)
2.group by方法
查數據:
select count(num), max(name) from student --列出重復的記錄數，并列出他的name屬性
group by num
having count(num) >1 --按num分組后找出表中num列重復，即出現次數大于一次
刪數據：
delete from student
group by num
having count(num) >1
這樣的話就把所有重復的都刪除了。
3.用distinct方法 -對于小的表比較有用
create table table_new as select distinct * from table1 minux
truncate table table1;
insert into table1 select * from table_new;
