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Question

我有如下表：

create table passenger (passenger_ID varchar (10) , passenger_name varchar (30) , passenger_city varchar (30) , primary key (passenger_ID)) ; 
    create table flight (flight_number varchar (10) , departure_airport varchar (10), arrival_airport varchar (10) , primary key (flight_number)); 

    create table seat ( 
    flight_number varchar (10), 
    seat_number varchar (10), 
    primary key(flight_number, seat_number), 
    foreign key(flight_number) references flight); 

    create table reservation(
  passenger_ID varchar (10) 
  flight_number varchar (10) , seat_number varchar (10) , 
    day date ,
  fare numeric(8,2), 
    primary key (flight_number, seat_number,day), 
    foreign key (flight_number, seat_number) references seat) , 
    foreign key (passenger_ID) references passenger ; 

我被困在了这个问题： 查找保留一个航班每位乘客的ID（或航班）从“ALB”出发，从未预订到达'ALB'的航班。对于此查询，请使用in子句或not in子句。

我写了这么多查询：

select a.passenger_id from 
reservation a 
where a.flight_number in (select b.flight_number from flight b where 
             b.departure_airport = 'ALB' 
            and b.arrival_airport <> 'ALB'); 

但这种查询是不正确的。子查询返回与过滤条件相匹配的所有航班号。但是主要的select查询会返回子查询返回的flight_number的所有乘客id，即使该乘客的arrival_airport是ALB。

我的预约表：

passenger_id flight_number seat_number day fare 
2 A2 201 10/1/17 1083 
1 A3 301 10/1/17 1173.25 
1 A4 402 10/1/17 846.81 
2 A5 501 10/1/17 752.72 
1 A5 502 10/1/17 485 
4 A5 506 10/1/17 970 
2 A6 601 10/1/17 388 
1 A7 703 10/1/17 921.5 
3 A7 704 10/1/17 921.5 
2 A8 804 10/1/17 970 
4 A8 805 10/1/17 970 

我的飞行表：

flight_number departure_airport arrival_airport 
A1 Ktm Pkr 
A2 Ktm NY 
A3 Ktm Ind 
A4 Ktm Chn 
A5 ALB KTM 
A6 ALB PKR 
A7 KTM ALB 
A8 PKR ALB 

Answer 1

想：两个条件。所以从你的思路来看：

select r.passenger_id 
from reservation r 
where r.flight_number in (select f.flight_number 
          from flight f 
          where f.departure_airport = 'ALB' 
         ) and 
     r.flight_number not in (select f.flight_number 
           from flight f 
           where f.arrival_airport = 'ALB' 
          ); 

但是，那不行。任何到达ALB的航班都将从其他机场出发。所以，这只是选择所有预订ALB航班的乘客。此逻辑在航班级别，而不是乘客级别。

嗯。您实际上需要的信息是所有航班客户“一次”。这表明聚合：

select r.passenger_id 
from reservation r join 
    flight f 
    on r.flight_number = f.flight_number 
group by r.passenger_id 
having sum(case when f.departure_airport = 'ALB' then 1 else 0 end) > 0 and -- reserved a flight from ALB 
     sum(case when f.arrival_airport = 'ALB' then 1 else 0 end) = 0;  -- never reserved a flight to ALB 

Answer 2

只是为了以稍微不同的方式证明@ GordonLinoff的答案的正确性，让我们稍微区别看待它。 :-)（See this SQLFiddle）。 （请注意，我填充了PASSENGER表，我们稍后会使用它。:-)。首先，让我们抓住所有谁已预留至少一个航班从ALB离开的乘客：

-- Find passengers who have reserved at least one flight departing from ALB 

SELECT DISTINCT r.PASSENGER_ID 
    FROM RESERVATION r 
    INNER JOIN FLIGHT f 
    ON f.FLIGHT_NUMBER = r.FLIGHT_NUMBER 
    WHERE f.DEPARTURE_AIRPORT = 'ALB'; 

这将返回PASSENGER_ID的1，2，4

接下来，让我们找到所有谁在保留乘客至少一个飞行到达ALB：

-- Find passengers who have reserved at least one flight arriving at ALB 

SELECT DISTINCT r.PASSENGER_ID 
    FROM RESERVATION r 
    INNER JOIN FLIGHT f 
    ON f.FLIGHT_NUMBER = r.FLIGHT_NUMBER 
    WHERE f.ARRIVAL_AIRPORT = 'ALB'; 

这返回PASSENGER_ID的1，2，3和4。

上述结果的琐碎审查表明显然，没有任何乘客预订了ALB航班，并且没有预订到达ALB的航班。但是因为我们只想让查询向我们展示我们想要什么，而不必去思考（毕竟，这是计算机应该做的事情:-)我们将把上述查询放在一起：

-- Now put them together 

SELECT p.PASSENGER_ID 
    FROM PASSENGER p 
    INNER JOIN (SELECT DISTINCT r.PASSENGER_ID 
       FROM RESERVATION r 
       INNER JOIN FLIGHT f 
        ON f.FLIGHT_NUMBER = r.FLIGHT_NUMBER 
       WHERE f.DEPARTURE_AIRPORT = 'ALB') d 
    ON d.PASSENGER_ID = p.PASSENGER_ID 
    LEFT OUTER JOIN (SELECT DISTINCT r.PASSENGER_ID 
        FROM RESERVATION r 
        INNER JOIN FLIGHT f 
         ON f.FLIGHT_NUMBER = r.FLIGHT_NUMBER 
        WHERE f.ARRIVAL_AIRPORT = 'ALB') a 
    ON a.PASSENGER_ID = p.PASSENGER_ID 
    WHERE a.PASSENGER_ID IS NULL; 

如预期的那样，此查询不返回任何结果。

但...我们真的想确保我们的查询真的有效。 Soooo ...让我们添加几行到几个表。首先，我们将添加一个新的乘客：

insert into passenger (passenger_ID, passenger_name, passenger_city) 
    values ('5', 'Bugs Bunny', 'ABC'); 

，我们还会增加对兔子先生，一个在那里，他离开ALB航班：

INSERT INTO RESERVATION (PASSENGER_ID, FLIGHT_NUMBER, SEAT_NUMBER, DAY, FARE) 
    VALUES ('5', 'A6', '123', TO_DATE('2017-10-01','YYYY-MM-DD'), 400); 

所以现在我们有乘员5，先生B. Bunny，离开ALB--但他从未飞入ALB。 （我认为那是因为他应该在阿尔伯克基左转弯:-)。所以我们的查询应该返回乘客5，并且if you go look at this SQLFiddle你会看到发生的情况。

Ehhh - 祝你好运，doc。

Answer 3

集中在乘客的替代航班，使用IN()和NOT IN()

SELECT DISTINCT 
     r.passenger_id 
FROM reservation r 
WHERE r.passenger_id IN (
     SELECT 
      r.passenger_id 
     FROM RESERVATION r 
     INNER JOIN FLIGHT f ON f.flight_number = r.flight_number 
     WHERE f.DEPARTURE_AIRPORT = 'ALB' 
    ) 
AND r.passenger_id NOT IN (
     SELECT 
      r.passenger_id 
     FROM RESERVATION r 
     INNER JOIN FLIGHT f ON f.flight_number = r.flight_number 
     WHERE f.ARRIVAL_AIRPORT = 'ALB' 
    ) 
; 

但是鲍勃贾维斯分析也表明，在您的样本数据中没有行符合分配条件。