MySQL删除重复的反向值

时间:2011-03-25 22:26:49

标签: php mysql myisam duplicate-removal

我有MySQL MyISAM表:

表朋友(id,friend_id):

1,5

5,1

2,6

6,2

3,7

如何删除反向记录?如果记录值“1,5”存在值为“5,1”的记录,我需要删除“5,1”。

Thanx寻求帮助!

4 个答案:

答案 0 :(得分:7)

DELETE F1
FROM friends F1, friends F2
WHERE F1.friend_id = F2.id
  AND F2.friend_id = F1.id
  AND F1.id > F1.friend_id

修改

更好的语法是:

DELETE F1
FROM friends F1
  JOIN friends F2 ON F1.friend_id = F2.id AND F2.friend_id = F1.id
WHERE F1.id > F1.friend_id

但执行时间是一样的。

同样,我已经快速创建了this small script&肮脏的基准。

<强>结果:

没有索引:

Dalen: 600 => 400 rows. Time: 0.0274
Mark: 600 => 400 rows. Time: 0.4323
Frosty: 600 => 400 rows. Time: 0.4081
Nick: 600 => 400 rows. Time: 0.3201

idfriend_id上的单独索引:

Dalen: 600 => 400 rows. Time: 0.0201
Mark: 600 => 400 rows. Time: 0.0095
Frosty: 600 => 400 rows. Time: 0.0059
Nick: 600 => 400 rows. Time: 0.3257

(id, friend_id)上的唯一索引:

Dalen: 600 => 400 rows. Time: 0.0168
Mark: 600 => 400 rows. Time: 0.0057
Frosty: 600 => 400 rows. Time: 0.0041
Nick: 600 => 400 rows. Time: 0.3209

结论:

  • Dalen:列未编入索引时速度最快
  • Frosty:当列被索引时最快(Mark在“索引情况”中接近甚至更快,而不考虑tmp表创建时间。但是,当字段被索引时,tmp表创建所需的额外时间会增加)

答案 1 :(得分:3)

试试这个:

create temporary table tmp    
  select a.* from friends as a,friends as b 
  where a.id = b.friend_id 
    and a.friend_id = b.id    /* left out of original post */
    and a.id < b.id;

delete from friends using friends inner join tmp 
   on friends.id = tmp.id
  and friends.friend_id=tmp.friend_id;

如果朋友表非常大,则联接比其他方法更有效。

编辑:我修复了上面的“创建...选择”语句。测试它,工作正常。对不起!

答案 2 :(得分:2)

DELETE FROM friends
WHERE (id,friend_id) IN
(   
   SELECT * FROM
   (
      SELECT t1.id,t1.friend_id
      FROM friends t1 JOIN friends t2
         ON t1.id=t2.friend_id AND t1.friend_id = t2.id
      WHERE t1.id > t1.friend_id
   ) t3
)

答案 3 :(得分:1)

create table friends (
id int,
friend_id int ) engine = myisam;

insert into friends values (1,5),(5,1),(2,6),(6,2),(3,7);


delete from friends where (id,friend_id) in
(select * from (
select id,friend_id from friends
union all
select friend_id,id from friends ) as t
where id > friend_id
group by id,friend_id
having count(*) >  1)