如何在postgres sql中连接3个表？

Question

我正在解决 Danny Ma 的 Pizza Runner SQL 案例研究。在“定价和评级”一节中，我陷入了两个问题，无法找到正确的解决方案。 案例研究链接：https://8weeksqlchallenge.com/case-study-2/

[cleaned_customer_orders table](https://i.stack.imgur.com/fEgXb.png)
[cleaned_runner_orders table](https://i.stack.imgur.com/sl6kd.png)
[extras table](https://i.stack.imgur.com/Ggm0B.png)

Q1。查询

select   
    sum(case when cco.pizza_id = 1 then 12 else 10 end) as pizza_cost  
from cleaned_customer_orders cco  
join pizza_names pn on pn.pizza_id = cco.pizza_id  
join cleaned_runner_orders cro on cro.order_id = cco.order_id  
where cro.cancellation is null  

Q2。查询

with cte_extras_cost as (select  
    sum(case when pn.pizza_id = 1 then 12 else 10 end) as pizza_cost,  
    count (e.topping_id)*1 as extras_cost  
    from cleaned_customer_orders cco  
    join pizza_names pn on pn.pizza_id = cco.pizza_id  
    left join extras e on e.record_id = cco.record_id  
    join cleaned_runner_orders cro on cro.order_id = cco.order_id  
    where cro.cancellation is null)  
select pizza_cost, coalesce(extras_cost, 0) as extras_cost, (pizza_cost+coalesce(extras_cost, 0))  
 as total_cost  
from cte_extras_cost;  

Q5。查询

with cte as (  
    select   
    sum(cro.distance*0.30) as delivery_cost,  
    sum(case when cco.pizza_id = 1 then 12 else 10 end) as total_cost  
from cleaned_runner_orders cro  
join cleaned_customer_orders cco on cro.order_id = cco.order_id  
where cro.cancellation is null)  
select cte.total_cost, cte.delivery_cost, (cte.total_cost - cte.delivery_cost) as actual_cost  
from cte  

我的输出：
Q1. 138
Q2。披萨成本=150，额外成本=4，总成本=154
Q5.总成本=138，交货成本=64，实际成本=73

正确答案是
Q1. 138
Q2。披萨成本=138，额外成本=4，总成本=142
Q5.总成本=138，交货成本=45，实际成本=73

Answer 1

如果您发布插入语句来填充临时表会更好

q1

select   
    sum(case when cco.pizza_id = 1 then 12 else 10 end) as pizza_cost  
from customer_orders cco  
left join runner_orders cro on cro.order_id = cco.order_id  
where coalesce(cro.cancellation,'') not in ('Restaurant Cancellation', 'Customer Cancellation')

q2

with 
cco as 
(select *, row_number() over(order by order_id, pizza_id) as rn from customer_orders)
, ex as (
select rn, case when extras = 'null' then null else extras end as extras from 
            cco
            where coalesce(case when extras = 'null' then null else extras end,'') <> ''
)
, ex1 as (
select trim(unnest(string_to_array(extras, ',')))::int as topping_id, rn from ex
)
, extras as (
select count(topping_id) as topping_cnt, rn from ex1 group by rn
)
, cte_extras_cost as (
    select 
    sum(case when pn.pizza_id = 1 then 12 else 10 end) pizza_cost
    , sum(topping_cnt*1) as extras_cost
    from cco  
    join pizza_names pn on pn.pizza_id = cco.pizza_id  
    left join extras e on e.rn = cco.rn  
    join runner_orders cro on cro.order_id = cco.order_id  
    where coalesce(cro.cancellation,'') not in ('Restaurant Cancellation', 'Customer Cancellation')
)  
select pizza_cost, coalesce(extras_cost, 0) as extras_cost, (pizza_cost+coalesce(extras_cost, 0))  
 as total_cost  
from cte_extras_cost;

q5

with  cro as  (     select          case when distance <> '' then cast(distance as numeric) else 0 end as distance, order_id, runner_id,        pickup_time, duration, cancellation     from (  SELECT order_id, runner_id, pickup_time, replace(replace(distance, 'km', ''), 'null', '') as distance,      duration, cancellation      from    pizza_runner.runner_orders  ) q1 ) , cco as ( select sum(case when pizza_id = 1 then 12 else 10 end) as pizza_cost, order_id from customer_orders group by order_id  ) , cte as (  
    select   
    sum(cro.distance*0.30) as delivery_cost, 
    sum(pizza_cost) as total_cost from cro   join cco on cro.order_id = cco.order_id   where coalesce(cro.cancellation,'') not in ('Restaurant Cancellation', 'Customer Cancellation') )   select  cte.total_cost, cte.delivery_cost, (cte.total_cost - cte.delivery_cost) as actual_cost   from cte

q1 和 q2 有明确的决定，我无法找到如何从这些数字中得到 45 的运费