MySql数据库：数据查重、去重的实现

数据库的查重去重是java面试经常会被问到的问题，下面讲讲怎么实现这些，因为本人能力有限，所以只管实现，不考虑效率问题；

假设有一个表user，字段分别有id–nick_name–password–email–phone，分情况如下（注意删除多余记录时要创建临时表，不然会报错）：

单字段（nick_name）

查出所有有重复记录的所有记录

select * from user where nick_name in
     (select nick_name from user group by nick_name having count(nick_name)>1);

查出有重复记录的各个记录组中id最大的记录

select * from user where id in (select max(id) from user group by nick_name having count(nick_name)>1);

查出多余的记录，不查出id最小的记录

select * from user where nick_name in

(select nick_name from user group by nick_name having count(nick_name)>1)

and id not in

(select min(id) from user group by nick_name having count(nick_name)>1);

删除多余的重复记录，只保留id最小的记录

delete from user where nick_name in
     (select nick_name from

(select nick_name from user group by nick_name having count(nick_name)>1) as tmp1)

and id not in

(select id from

(select min(id) from user group by nick_name having count(nick_name)>1) as tmp2);

多字段（nick_name,password）

查出所有有重复记录的记录

select * from user where (nick_name,password) in

(select nick_name,password from user group by nick_name,password where having count(nick_name)>1);

查出有重复记录的各个记录组中id最大的记录

select * from user where id in

(select max(id) from user group by nick_name,password where having count(nick_name)>1);

查出各个重复记录组中多余的记录数据，不查出id最小的一条

select * from user where (nick_name,password) in

(select nick_name,password from user group by nick_name,password having count(nick_name)>1)

and id not in

(select min(id) from user group by nick_name,password having count(nick_name)>1);

删除多余的重复记录，只保留id最小的记录

delete from user where (nick_name,password) in

(select nick_name,password from

(select nick_name,password from user group by nick_name,password having count(nick_name)>1) as tmp1)

and id not in

(select id from

(select min(id) id from user group by nick_name,password having count(nick_name)>1) as tmp2);

作者：zilie972803341
来源：优快云
原文：https://blog.youkuaiyun.com/zilie972803341/article/details/78444703
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