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完美的【去重留一】SQL

2018年12月07日  | 移动技术网IT编程  | 我要评论

 

delete consum_record
from
 consum_record, 
 (
  select
   min(id) id,
   user_id,
   monetary,
   consume_time
  from
   consum_record
  group by
   user_id,
   monetary,
   consume_time
  having
   count(*) > 1
 ) t2
where
 consum_record.user_id = t2.user_id 
 and consum_record.monetary = t2.monetary
 and consum_record.consume_time = t2.consume_time
and consum_record.id > t2.id;

上面这条sql语句，仔细看一下，揣摩出思路也不难，大概也分为3步来理解：

(select min(id) id, user_id, monetary, consume_time from consum_record group by user_id, monetary, consume_time having count(*) > 1 ) t2 查询出重复记录形成一个集合（临时表t2），集合里是每种重复记录的最小id
consum_record.user_id = t2.user_id and consum_record.monetary = t2.monetary and consum_record.consume_time = t2.consume_time 关联 判断重复基准的字段

根据条件，删除原表中id大于t2中id的记录

 

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