Question

我不知道问题的标题是否足够清楚，但我有这种情况：

TABLE ID, VALUE_1, VALUE_2
       1, HORSE  , 500
       1, DOG    , 400
       1, DUCK   , 300
       2, HORSE  , 500
       2, DOG    , 400
       2, DUCK   , 300

我想查看

之类的值

ID HORSE DOG DUCK
1    500 400 300
2    500 400 300

Answer 1

您可以使用条件聚合：

select id,
       sum(case when value_1 = 'HORSE' then value_2 end) as horse,
       sum(case when value_1 = 'DOG' then value_2 end) as dog,
       sum(case when value_1 = 'DUCK' then value_2 end) as duck
from t
group by id
order by id;

Answer 2

使用condititonal aggregation

select id,
       max(case when value_1 = 'HORSE' then value_2 end ) as horse,
       max(case when value_1 = 'DOG' then value_2 end ) as dog,
       max(case when value_1 = 'DUCK' then value_2 end ) as duck
  from tab
 group by id;

Demo

Answer 3

有条件聚合：

select
  ID,
  max(case when value_1 = 'HORSE' THEN value_2 end) HORSE,
  max(case when value_1 = 'DOG' THEN value_2 end) DOG,
  max(case when value_1 = 'DUCK' THEN value_2 end) DUCK
from tablename
group by ID

Answer 4

在Oracle 11.1和更高版本中，可以使用PIVOT运算符。假设表名称为T：

select id, horse, dog, duck
from   t
pivot  (min(value_2) 
        for value_1 in ('HORSE' as horse, 'DOG' as dog, 'DUCK' as duck))
order  by id
;

在Oracle 12.1和更高版本中，可以使用MATCH_RECOGNIZE实现相同的功能（并且我们不再需要聚合）：

select id, horse, dog, duck
from   t
match_recognize(
  partition by id
  measures  horse.value_2 as horse,
            dog.value_2   as dog,
            duck.value_2  as duck
  pattern   ( (horse|dog|duck)* )
  define    horse as value_1 = 'HORSE', dog as value_1 = 'DOG', 
             duck as value_1 = 'DUCK'
);

使用选择值进入列

4 个答案: