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  • mysql 去重

    https://blog.csdn.net/eagle89/article/details/90901755

    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):

    一、单字段(nick_name)

    1、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);

    2、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);

    3、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);

    4、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);

    二、多字段(nick_name,password)

    1、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);

    2、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);

    3、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);

    4、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);

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  • 原文地址:https://www.cnblogs.com/zonglonglong/p/13678504.html
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