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oracle sql 去重复记录不用distinct如何实现

Jun 07, 2016 pm 05:55 PM
distinct oracle

本文将详细介绍oracle sql 去重复记录不用distinct如何实现,需要了解的朋友可以参考下

用distinct关键字只能过滤查询字段中所有记录相同的(记录集相同),而如果要指定一个字段却没有效果,另外distinct关键字会排序,效率很低 。
select distinct name from t1 能消除重复记录,但只能取一个字段,现在要同时取id,name这2个字段的值。
select distinct id,name from t1 可以取多个字段,但只能消除这2个字段值全部相同的记录
所以用distinct达不到想要的效果,用group by 可以解决这个问题。
例如要显示的字段为A、B、C三个,而A字段的内容不能重复可以用下面的语句:
select A, min(B),min(C),count(*) from [table] where [条件] group by A
having [条件] order by A desc
为了显示标题头好看点可以把select A, min(B),min(C),count(*) 换称select A as A, min(B) as B,min(C) as C,count(*) as 重复次数
显示出来的字段和排序字段都要包括在group by 中
但显示出来的字段包有min,max,count,avg,sum等聚合函数时可以不在group by 中
如上句的min(B),min(C),count(*)
一般条件写在where 后面
有聚合函数的条件写在having 后面
如果在上句中having加 count(*)>1 就可以查出记录A的重复次数大于1的记录
如果在上句中having加 count(*)>2 就可以查出记录A的重复次数大于2的记录
如果在上句中having加 count(*)>=1 就可以查出所有的记录,但重复的只显示一条,并且后面有显示重复的次数----这就是所需要的结果,而且语句可以通过hibernate
下面语句可以查询出那些数据是重复的:
select 字段1,字段2,count(*) from 表名 group by 字段1,字段2 having count(*) > 1
将上面的>号改为=号就可以查询出没有重复的数据了。
例如
代码如下:
select count(*) from (select gcmc,gkrq,count(*) from gczbxx_zhao t group by gcmc,gkrq having
count(*)>=1 order by GKRQ)
select * from gczbxx_zhao where viewid in ( select max(viewid) from gczbxx_zhao group by
gcmc ) order by gkrq desc ---还是这个可行

有一面试题说:distinct去重复的效率很底下,我在网上看到这遍文章的方法好像说是用 group by having 效率很高了?
我在了一个测试,有一商品表,26万条记录,只有商品编号建了索引,对品牌名称字段做 distinct
代码如下:
select brand,count(*) from tab_commbaseinfo group by brand having count(*) =1

平均时间是:0.453
代码如下:
select distinct brand from tab_commbaseinfo

平均时间是:0.39
搞不懂是否还有其它方法。
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