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 程式師世界 >> 數據庫知識 >> 其他數據庫知識 >> 更多數據庫知識 >> 查詢及刪除重復記錄

查詢及刪除重復記錄

編輯:更多數據庫知識

(一)
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,*,Count(*) From A Group By Name,* 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、對於第一種重復,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)
  最後一個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
)
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)
運行會產生問題,where(a.peopleId,a.seq)這樣的寫發是通不過的!!!

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