Question

我想计算值在某些列中出现的频率并创建一个新表，其中值为列，频率为数据。例如：

create table users
(id number primary key, 
name varchar2(255));

insert into users  values (1, 'John');
insert into users  values (2, 'Joe');
insert into users  values (3, 'Max');

create table meals
(id number primary key,
user_id number,
food varchar2(255));

insert into meals values (1, 1, 'Apple');
insert into meals values (2, 1, 'Apple');
insert into meals values (3, 1, 'Orange');
insert into meals values (4, 1, 'Bread');
insert into meals values (5, 1, 'Apple');
insert into meals values (6, 2, 'Apple');
insert into meals values (7, 2, 'Bread');
insert into meals values (8, 2, 'Bread');
insert into meals values (9, 2, 'Apple');
insert into meals values (10, 3, 'Orange');
insert into meals values (11, 3, 'Bread');
insert into meals values (12, 3, 'Bread');

所以我得到了不同的用户和他们的饭菜（这里是面包，苹果和橘子）。对于每个用户，我想知道他多久吃一次不同的食物。以下查询完全符合我的要求：

select 
(select count(id) from meals where meals.user_id = users.id and meals.food = 'Apple') as count_apple,
(select count(id) from meals where meals.user_id = users.id and meals.food = 'Orange') as count_orange,
(select count(id) from meals where meals.user_id = users.id and meals.food = 'Bread') as count_bread
from users;

问题是，这真的很慢，特别是当我有超过100,000个用户和几十种不同的食物时。我确信有一种更快的方法，但我在SQL中没有足够的经验来解决这个问题。

Answer 1

如果您使用的是11g，则可以使用pivot运算符，如下所示：

select * from (
  select user_id, food from meals
)
pivot (count(*) as count for (food) in ('Apple', 'Orange', 'Bread'));

否则你必须做一个手动支点：

select user_id, 
       sum(case when food = 'Apple' then 1 else 0 end) count_apple, 
       sum(case when food = 'Orange' then 1 else 0 end) count_orange, 
       sum(case when food = 'Bread' then 1 else 0 end) count_bread
from   meals
group  by user_id

在任何一种情况下，这些都应该比原始方法更快，因为您只需访问meals表一次。

Oracle SQL：计算频率并转换为列

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