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發布時間:2023/12/18 编程问答 29 豆豆
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--如何按字段刪除重復記錄

一張表里面以兩個字段為唯一字段,當幾條記錄的這兩個字段完全相同時,需要刪除重復項,如下表
a b c d
1 2 3 4
1 5 3 5
1 2 7 9
以a、b為唯一字段,第一條和第三條的a、b完全相同,所以,需要刪除第一條記錄1
2 3 4 或者第三條記錄1 2 7 9
即如下結果:
a b c d
1 2 3 4
1 5 3 5

a b c d
1 5 3 5
1 2 7 9



CREATE TABLE Tb1(id int, [a] varchar(255), [b] varchar(255), [c] varchar(255), [d] varchar(255))
INSERT Tb1(id, [a], [b], [c], [d])
??????????
SELECT 1, '1','2','3','4'
UNION ALL? SELECT 2, '1','5','3','5'
UNION ALL? SELECT 3, '1','2','7','9'
UNION ALL? SELECT 4, '1','4','7','6'

delete Tb1 where [id] not in (select max([id]) from Tb1 group by a,b )
select * from tb1

drop table tb1

如果要同時刪除第一和第三行
即如下結果:
a b c d
1 5 3 5

語句如下:

delete m from tb t
inner join
(
select a ,b
from tb
group by a , b
having count(*)>1
)n
on m.a = n.a and m.b = n.b

delete * from tb as m,
(
select a ,b
from tb
group by a , b
having count(*)>1
)n
where m.a = n.a and m.b = n.b








------------------------------------------------------------------------------------
在幾千條記錄里,存在著些相同的記錄,用SQL語句,刪除掉重復記錄
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)

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

轉載于:https://www.cnblogs.com/zhuawang/archive/2008/07/15/1243369.html

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