Question

我有三张桌子：

BookingNode，预订AirTrip

AirTrip：

+----+------------+
| ID |    Name    |
+----+------------+
|  0 | One way    |
|  1 | Round trip |
|  2 | Circle     |
|  3 | Other      |
+----+------------+

我们在预订时将数据存储为： BookingNode表

+--------+-------------------+------------+----------------------+
|   ID   | CustomerGivenName | IPAddress  |        Email         |
+--------+-------------------+------------+----------------------+
| 177022 | xfghfh            | 2130706473 | mikehussey@gmail.com |
| 177021 | cfggjfj           | 2130706473 | mikehussey@gmail.com |
+--------+-------------------+------------+----------------------+

预订表：

+--------+---------------+-----------+------------+------------+
|   ID   | BookingNodeID | AirTripID | AirLineId  | Provider   |
+--------+---------------+-----------+------------+------------+
| 181251 |        177020 |         1 |        978 |        Jet |
| 181252 |        177021 |         0 |        982 |         Go |
| 181253 |        177021 |         0 |        978 |        Jet |
+--------+---------------+-----------+------------+------------+

如果预订往返航班且ProviderID相同，则在预订表中完成单个条目，AirTripID值为1.（预订ID：181251和提供者Jet）但如果提供商的两条腿都不同，则两个条目在预订表中使用AirTripID完成，两个条目都是一个（预订ID：181252和181253 Provider Go，Jet）。在这种情况下，BookingNodeID值相同。

问题：我必须编写一个查询来获取不同类型的预订。（Oneway，RoundTrip，Circle）。但是当我在AirTripID上应用联接时，它给了我不正确的结果。我如何编写我的查询以获得正确的结果，因为知道BookingNodeID对于往返行程是相同的（预订表中的两个条目）

示例输出

+-------------+---------------+-------------------+------------+
| AirTripName | BookingNodeID | CustomerGivenName | IPAddress  |
+-------------+---------------+-------------------+------------+
| TwoWay      |        177020 | xfghfh            | 2130706473 |
| TwoWay      |        177021 | cfggjfj           | 2130706473 |
+-------------+---------------+-------------------+------------+

Answer 1

基本上，由于我的数据输入懒惰，这段代码可能会出错。但是，查询的逻辑是，如果b.AirTripID为0，则通过Booking添加额外条件。如果结果返回超过1行，实际上是2路。所以AirTripType将变为1，否则，保持与b.AirTripID相同。您可以在下面复制并尝试修复，如果有任何错误。我相信逻辑应该基于你的预期结果。

  select 
  bd.ID,
  bd.CustomerGivenName,
  case b.AirTripID 
  when 1 then 1
  when 2 then 2
  when 3 then 3
  when 0 then
    case select BookingNodeID 
    from Booking 
    where Booking.BookingNodeID = bd.ID group by BookingNodeID having Count(BookingNodeID) 
    when 1 then 1 
    else 0 end as AirTripType,
  bd.IPAddress
  from BookingNode bd
  inner join (select BookingNodeID ,AirTripID  from Booking group by BookingNodeID ,AirTripID) as b on b.BookingNodeID = bd.ID
  where id=177021

Answer 2

试试这个

WITH CTE
AS
(
SELECT
    SeqNo = ROW_NUMBER() OVER(PARTITION BY BN.ID ORDER BY B.ID),
    B.BookingNodeID,
    BN.CustomerGivenName,
    BN.IPAddress,
    AirTripId = A.ID,
    AirTripNm = A.Name
    FROM Booking B
      INNER JOIN AirTrip A
        ON A.ID = B.AirTripID
      LEFT JOIN BookingNode BN
        ON B.BookingNodeID = BN.id
)
SELECT
  C1.SeqNo,
  AirTripName = CASE WHEN C2.SeqNo IS NOT NULL
                        THEN 'Round trip'
                     ELSE C1.AirTripNm END,
  C1.BookingNodeID,
  C1.CustomerGivenName,
  C1.IPAddress
  FROM CTE C1
    LEFT JOIN CTE C2
      ON C1.BookingNodeID = C2.BookingNodeID
        AND C2.SeqNo = 2
    WHERE c1.SeqNo = 1

SQL Fiddle Link Here

Answer 3

Select  distinct bk.bookingnodeid,cst.customername,ipaddress,
case when count(airtripid)over(partition by bookingnodeid order by bookingnodeid)=2  then 'RoundTrip' else name end As AirTripName
from booking bk
inner join airlinetrip at
on bk.airtripid=at.id
inner join customer cst
on cst.id=bk.bookingnodeid

SQL查询操作

3 个答案: