请求协助（postgresql）最小/最大功能

我是博士后新生，提出了以下问题。我需要回答的问题是，是否有城市的经济价格比商业价格便宜。目前我的请求如下：

with cte_money as (
    select  
        ad.city ->> 'ru' as city,
        fare_conditions, 
        amount
    from ticket_flights tf   
    join flights as f using(flight_id)
    join airports_data as ad on f.arrival_airport = ad.airport_code 
    order by city, fare_conditions asc
    )
    select  
        city,
        (select max(amount) where fare_conditions = 'Economy') as Max_Economy,
        (select min(amount) where fare_conditions = 'Business') as Min_Business
    from cte_money
    group by city, fare_conditions
    order by city asc 

它给出以下黄色结果：

---

如何将同一城市的最大值与最小值合并到一行中，并相互比较？总之，我只需要显示经济舱比商务舱便宜的城市。提前感谢您

对于此查询分组，当您将条件放入聚合中时，无需进行子选择。然后，您可以使用having子句轻松筛选更便宜的经济舱：

with cte_money as (
    select  
        ad.city ->> 'ru' as city,
        fare_conditions, 
        amount
    from ticket_flights tf   
    join flights as f using(flight_id)
    join airports_data as ad on f.arrival_airport = ad.airport_code 
    order by city, fare_conditions asc
    )
    select  
        city,
        max(CASE WHEN fare_conditions = 'Economy' THEN amount ELSE NULL END) as Max_Economy,
        min(CASE WHEN fare_conditions = 'Business' THEN amount ELSE NULL END) as Min_Business 
    from cte_money
    group by city, fare_conditions
    HAVING max(CASE WHEN fare_conditions = 'Economy' THEN amount ELSE NULL END) <= min(CASE WHEN fare_conditions = 'Business' THEN amount ELSE NULL END)
    order by city asc 

---

我理解您的问题和模式，不需要CTE或嵌套查询。您可以使用聚合和having子句进行筛选：

select ad.city ->> 'ru' as city,
from ticket_flights tf   
join flights as f using(flight_id)
join airports_data as ad on f.arrival_airport = ad.airport_code 
group by ad.city ->> 'ru'
having 
    max(amount) filter (where fare_conditions = 'Economy') 
    < min(amount) filter(where fare_conditions = 'Business') 
order by city

如果要在结果集中显示条件最小值和最大值，可以在select子句中重复它们