Question

QUOTES TABLE
-----------------
QUOTE_ID_PK  PICKUP_FROM_ID  DROPOFF_TO_ID
1               4                     3
2               2                     3
3               1                     1
4               1                     2
5               3                     2


TOWN TABLE
---------------
 TOWN_ID_PK  DISTANCE    NAME
    1        34          South Central 
    2        43          airport
    3        99          bus station
    4        66          train station

我有2个表来计算一个简单的距离公式。

我需要在订单表中为ORDER_ID返回2行（带距离值）。

我得到order_id = 2的结果，pickupfromdistance = 43，dropoffto = 99。

我已经尝试过这个网站的一些想法，这就是我所拥有的几乎可以工作，但不会返回第二个距离值。我是否以错误的方式解决了这个问题？我是否为一个简单的问题提出了复杂的解决方案？

SELECT town.town_id_pk AS townID, town.distance AS distance, town.name AS townName, town.location AS location, quotes.pickupfrom_id, quotes.dropoffto_id 
FROM town, quotes 
WHERE quotes.pickupfrom_id=town.town_id_pk 
AND quotes.quote_id_pk = '140419420'
UNION ALL
SELECT town.town_id_pk AS townID, town.distance AS distance, town.name AS townName, town.location AS location, quotes.pickupfrom_id, quotes.dropoffto_id
FROM town, quotes 
WHERE quotes.dropoffto_id=town.town_id_pk 
AND quotes.quote_id_pk = '140419420'";

E.G。我在选择：

QUOTE_ID_PK=2, wh

（PICKUP_FROM_ID = 2，DROPOFF_TO_ID = 3）

我所追求的结果将是这样的：

TOWN_ID_PK=2, DISTANCE=43, NAME=AIRPORT

然后下一个ROW将是

TOWN_ID_PK=3, DISTANCE=99, NAME= BUS STATION

一旦我知道哪个距离更远，我就可以继续使用代码来计算每千克的速率等等（我可以自己完成这部分）。

Answer 1

只是给出一个提示：

SELECT 
  ORDER_ID, 
  PICKUP_FROM_TOWN_ID, 
  (SELECT DISTANCE FROM TOWN  where TOWN_ID_PK = quotes.DROPOFF_TO_ID) as DISTANCE, 
  quotes.DROPOFF_TO_ID  
from town, quotes 
WHERE quotes.PICKUP_FROM_TOWN_ID=town.TOWN_ID_PK
AND ORDER_ID = '140419420'

这是一个sqlfiddle

<强>编辑：

为了得到两行，试试这个：

SELECT 
  ORDER_ID, 
  PICKUP_FROM_TOWN_ID, 
  DISTANCE,
  DROPOFF_TO_ID  
from town, quotes 
WHERE 
    ( quotes.PICKUP_FROM_TOWN_ID=town.TOWN_ID_PK or quotes.DROPOFF_TO_ID=town.TOWN_ID_PK )
AND ORDER_ID = 2

新sqlfiddle

Answer 2

您是否只是为了争取两个城镇的距离？如果是这样，您可以使用join执行此操作：

select oq.*, tfrom.distance as distance_from, tto.distance as to_distance
from OrderQuotes oq left outer join
     Towns tfrom
     on oq.PICKUP_FROM_TOWN_ID = tfrom.town_id_pk left outer join
     Towns tto
     on oq.PICKUP_TO_TOWN_ID = tto.town_id_pk;

使用2个表PHP从一个SELECT sql查询中检索2行

2 个答案: