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我正在构建一个口袋妖怪数据集，并希望对其运行一些查询。这是数据库的设置：

create table pokedex(
    name varchar(20) not null,
    weigth int not null,
    height int not null,
    primary key(name)
);
create table trainer(
    name varchar(20) not null,
    location varchar(20) not null,
    gender varchar(10) not null,
    birth_year int not null,
    primary key(name)
);

create table trainer_pokemon(
    trainer_name varchar(20) not null,
    pokemon_name varchar(20) not null,
    level int not null,
    year_obtained int not null,
    primary key(trainer_name, pokemon_name, level, year_obtained),
    foreign key(trainer_name) references trainer(name),
    foreign key(pokemon_name) references pokedex(name)
);
create table type(
    name varchar(20) not null,
    primary key(name)
);
create table poke_type(
    pokemon_name varchar(20) not null,
    type_name varchar(20) not null,
    primary key(pokemon_name, type_name),
    foreign key(pokemon_name) references pokedex(name),
    foreign key(type_name) references type(name)
);

这个想法是数据集不应该有冗余数据，所以如果我想为每个训练师获取一个包含最常用口袋妖怪类型的表，我需要为每种类型获取一个表，或者至少这是我认为 atm 的表：

with psychics as (
    select trainer_name, count(type_name) psychic from trainer_pokemon as tp
    inner join poke_type as pt on pt.pokemon_name = tp.pokemon_name
    group by tp.trainer_name, pt.type_name
    having pt.type_name = 'Psychic'
),
waters as (
    select trainer_name, count(type_name) water from trainer_pokemon as tp
    inner join poke_type as pt on pt.pokemon_name = tp.pokemon_name
    group by tp.trainer_name, pt.type_name
    having pt.type_name = 'Water'
),
select tp.trainer_name, w.water, p.psychic from trainer_pokemon as tp
inner join waters as w on w.trainer_name = tp.trainer_name
inner join psychics as p on p.trainer_name = tp.trainer_name
group by tp.trainer_name, w.water, p.psychic

但是，这不会导致训练师没有特定类型的口袋妖怪（本例中为水/通灵）。

有没有人能指出我正确的方向来建立训练师表，其中特定类型在他们的口袋妖怪系列中出现的次数？

4

1 回答 1

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假设您使用的是相对更新的 Postgresql 版本（您已经标记了 Mysql 和 Postgres，它们是完全不同的数据库）

SELECT tp.trainer_name,
  COUNT(*) FILTER (WHERE pt.type_name = 'Psychic') as psychic,
  COUNT(*) FILTER (WHERE pt.type_name = 'Water')   as water
FROM trainer_pokemon tp, poke_type pt
WHERE tp.pokemon_name = pt.pokemon_name
GROUP BY 1

在您的问题中发布的查询可能会作为对表的多次扫描执行，但如果您确实想要这样做，您想使用“LEFT OUTER JOIN”（OUTER 关键字是可选的）：

with psychics as (
    select trainer_name, count(type_name) AS psychic
    from trainer_pokemon as tp
      inner join poke_type as pt on pt.pokemon_name = tp.pokemon_name
    group by tp.trainer_name, pt.type_name
    having pt.type_name = 'Psychic'
), waters as (
    select trainer_name, count(type_name) AS water
    from trainer_pokemon as tp
      inner join poke_type as pt on pt.pokemon_name = tp.pokemon_name
    group by tp.trainer_name, pt.type_name
    having pt.type_name = 'Water'
)
select tp.trainer_name, w.water, p.psychic
from trainer_pokemon as tp
  left join waters as w on w.trainer_name = tp.trainer_name
  left join psychics as p on p.trainer_name = tp.trainer_name

或者如果您获得两种以上的类型，则可能会更好地扩展其他方式：

with trained as (
    select trainer_name, pt.type_name, count(*) as num
    from trainer_pokemon as tp
      inner join poke_type as pt on pt.pokemon_name = tp.pokemon_name
    group by tp.trainer_name, pt.type_name
)
select tp.trainer_name, w.num as water, p.num as psychic
from trainer_pokemon as tp
  left join trained as w on tp.trainer_name = w.trainer_name and w.type_name = 'Water'
  left join trained as p on tp.trainer_name = p.trainer_name and p.type_name = 'Psychic'
于 2018-11-07T17:19:31.493 回答