Mysql刪除重復記錄,保留id最小的一條


mysql 查詢重復字段,及刪除重復記錄的方法
MySQL, 數據庫, 數據庫, 字段, 服務器
數據庫中有個大表,需要查找其中的名字有重復的記錄id,以便比較。如果僅僅是查找數據庫中name不重復的字段,很容易:
SELECT min(`id`),`name` FROM `table` GROUP BY `name`;

但是這樣並不能得到說有重復字段的id值。(只得到了最小的一個id值)查詢哪些字段是重復的也容易:
SELECT `name`,count(`name`) as count FROM `table` GROUP BY `name` HAVING count(`name`) >1 ORDER BY count DESC;

但是要一次查詢到重復字段的id值,就必須使用子查詢了,於是使用下面的語句。
SELECT `id`,`name` FROM `table` WHERE `name` in (
SELECT `name`
FROM `table`
GROUP BY `name` HAVING count(`name`) >1);

但是這條語句在mysql中效率太差,感覺mysql並沒有為子查詢生成零時表。於是使用先建立零時表:
create table `tmptable` as (
SELECT `name`
FROM `table`
GROUP BY `name` HAVING count(`name`) >1);

然后使用多表連接查詢:
SELECT a.`id`, a.`name` FROM `table` a, `tmptable` t WHERE a.`name` = t.`name`;

結果這次結果很快就出來了。
========================
查詢及刪除重復記錄的方法
(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)

(上面這條語句在mysql中執行會報錯:


執行報錯:1093 - You can't specify target table 'student' for update in FROM clause
原因是:更新數據時使用了查詢,而查詢的數據又做了更新的條件,mysql不支持這種方式。oracel和msserver都支持這種方
怎么規避這個問題?
再加一層封裝,

delete from php_user where 
username in (select username from ( select username from php_user group by username having count(username)>1) a) 
and id not in ( select min(id) from (select min(id) as id from php_user group by username having count(username)>1 ) b)

 注意select min(id) 后面要有as id.

 

 

 

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

  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,二是部分關鍵詞段重復的記錄,比如Name字段重復,而其它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

)

 

 

 

 

 

ref: http://www.greensoftcode.net/techntxt/2013120131251921893458
原文:http://bbs.ahpal.com/thread-3585-1-1.html


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