MySQL 删除数据库中重复数据方法小结

刚开始，根据我的想法，这个很简单嘛，上sql语句

delete from zqzrdp where tel in (select min(dpxx_id) from zqzrdp group by tel having count(tel)>1);

执行，报错！！~！~

异常意为：你不能指定目标表的更新在FROM子句。傻了，MySQL 这样写，不行，让人郁闷。

难倒只能分步操作，蛋疼

以下是网友写的，同样是坑爹的代码，我机器上运行不了。

1. 查询需要删除的记录，会保留一条记录。

select a.id,a.subject,a.RECEIVER from test1 a left join (select c.subject,c.RECEIVER,max(c.id) as bid from test1 c where status=0 GROUP BY RECEIVER,SUBJECT having count(1) >1) b on a.id< b.bid where a.subject=b.subject and a.RECEIVER = b.RECEIVER and a.id < b.bid

2. 删除重复记录，只保留一条记录。注意，subject,RECEIVER 要索引，否则会很慢的。

delete a from test1 a,(select c.subject,SUBJECT having count(1) >1) b where a.subject=b.subject and a.RECEIVER = b.RECEIVER and a.id < b.bid;

3. 查找表中多余的重复记录，重复记录是根据单个字段（peopleId）来判断

select * from people where peopleId in (select peopleId from people group by peopleId having count(peopleId) > 1)

4. 删除表中多余的重复记录，重复记录是根据单个字段（peopleId）来判断，只留有rowid最小的记录

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)

5.删除表中多余的重复记录（多个字段），只留有rowid最小的记录

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)

看来想偷懒使用一句命令完成这个事好像不太显示，还是老老实实的分步处理吧，思路先建立复制一个临时表，然后对比临时表内的数据，删除主表里的数据

alter table tableName add autoID int auto_increment not null; 

create table tmp select min(autoID) as autoID from tableName group by Name,Address; 

create table tmp2 select tableName.* from tableName,tmp where tableName.autoID = tmp.autoID; 

drop table tableName; 

rename table tmp2 to tableName;