Question

我要选择有条件的数据

我有一张桌子

Id|Name    |Age
1 |David   |1
2 |Aic     |2
3 |Owen    |2 
4 |Aic     |3
5 |Phuc    |3
6 |Aic     |4
7 |Ronaldo |4
8 |Ronaldo |5
9 |Ronaldo |6

我如何查询2条记录是否具有与2相同的年龄，它将只选择名称为“ aic”的记录，否则将获得全部（如果不相同的年龄将得到所有类似的图像）

Id|Name    |Age
1 |David   |1
2 |Aic     |2
4 |Aic     |3
6 |Aic     |4
8 |Ronaldo |5
9 |Ronaldo |6

Answer 1

您可以尝试使用row_number()

DEMO

select id,name,age from 
(
select id,name,age,row_number() over(partition by age order by name) as rn
from tablename
)A where rn=1

输出：

d   name    age
1   David   1
2   Aic     2
4   Aic     3
6   Aic     4
8   Ronaldo 5
9   Ronaldo 6

Answer 2

**您可以使用此** row_number()，而test1是您的表格名称

select id,name,age from  ( select id,name,age,row_number() 
        over(partition by age order by name) as rownumber from test1 )A where rownumber=1

Answer 3

一种简单的方法是：

select t.*
from t
where t.name = 'Aic'
union all
select t.*
from t
where t.name <> 'Aic' and
      not exists (select 1 from t t2 where t2.age = t.age and t2.name = 'Aic');

这将选择所有“ Aic”，然后选择所有其他没有Aic的年龄。实际上，这可以组合成一个查询：

select t.*
from t
where t.name = 'Aic' or
      (t.name <> 'Aic' and
       not exists (select 1 from t t2 where t2.age = t.age and t2.name = 'Aic')
      );

如果您想使用窗口功能，我可能会建议：

select t.*
from (select t.*,
             sum(case when name = 'Aic' then 1 else 0 end) over (partition by age) as num_aic
      from t
     ) t
where name = 'Aic' or num_aic = 0

Answer 4

随着您更改问题，请使用子查询

   select * from table where age in ( select age from table where name ='Aic')

当您再次更改问题时，它就会变成

with cte as (
         select *,
    row_number() over(partition by age order by  name) as rn
     from table
        ) select * from cte where rn=1

如果列中的相同数据将获得有条件的数据，如何选择数据

4 个答案: