oracle去除重复数据的方法:1、针对指定列,查出所有重复的行,并删除,方法为count having;2、删除所有重复的行,代码为【delete from nayi224_180824 t where t.rowid in】。 本文操作环境:Win
oracle去除重复数据的方法:1、针对指定列,查出所有重复的行,并删除,方法为count having;2、删除所有重复的行,代码为【delete from nayi224_180824 t where t.rowid in】。
本文操作环境:Windows7系统,oracle9i版本,Dell G3电脑。
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oracle去除重复数据的方法:
创建测试数据
create table nayi224_180824(col_1 varchar2(10), col_2 varchar2(10), col_3 varchar2(10));
insert into nayi224_180824select 1, 2, 3 from dual union allselect 1, 2, 3 from dual union allselect 5, 2, 3 from dual union allselect 10, 20, 30 from dual ;commit;select*from nayi224_180824;COL_1COL_2COL_3123123523102030
针对指定列,查出去重后的结果集
distinct
select distinct t1.* from nayi224_180824 t1;COL_1COL_2COL_3102030123523
方法局限性很大,因为它只能对全部查询的列做去重。如果我想对col_2,col3去重,那我的结果集中就只能有col_2,col_3列,而不能有col_1列。
select distinct t1.col_2, col_3 from nayi224_180824 t1COL_2COL_3232030
不过它也是最简单易懂的写法。
row_number()
select * from (select t1.*,
row_number() over(partition by t1.col_2, t1.col_3 order by 1) rn
from nayi224_180824 t1) t1 where t1.rn = 1;COL_1COL_2COL_3RN12311020301
写法上要麻烦不少,但是有更大的灵活性。
针对指定列,查出所有重复的行
count having
select * from nayi224_180824 t
where (t.col_2, t.col_3) in (select t1.col_2, t1.col_3
from nayi224_180824 t1
group by t1.col_2, t1.col_3
having count(1) > 1)COL_1COL_2COL_3123123523
要查两次表,效率会比较低。不推荐。
count over
select * from (select t1.*,
count(1) over(partition by t1.col_2, t1.col_3) rn
from nayi224_180824 t1) t1 where t1.rn > 1;COL_1COL_2COL_3RN123312335233
只需要查一次表,推荐。
删除所有重复的行
delete from nayi224_180824 t where t.rowid in (
select rid
from (select t1.rowid rid,
count(1) over(partition by t1.col_2, t1.col_3) rn
from nayi224_180824 t1) t1
where t1.rn > 1);
就是上面的语句稍作修改。
删除重复数据并保留一条
分析函数法
delete from nayi224_180824 t where t.rowid in (select rid
from (select t1.rowid rid,
row_number() over(partition by t1.col_2, t1.col_3 order by 1) rn
from nayi224_180824 t1) t1
where t1.rn > 1);
拥有分析函数一贯的灵活性高的特点。可以为所欲为的分组,并通过改变orderby从句来达到像”保留最大id“这样的要求。
group by
delete from nayi224_180824 t where t.rowid not in
(select max(rowid) from nayi224_180824 t1 group by t1.col_2, t1.col_3);
牺牲了一部分灵活性,换来了更高的效率。
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