Question

我一直在与客户合作，并继续回到这个查询问题，试图找到解决方案，我一直在谷歌搜索，但没有找到相关的结果。

我的客户有多种旅行服务，如游轮，海滩度假村，酒店，小木屋等。

我已经更新了所有单独的搜索页面，并且一直致力于让合并后的搜索页面正常运行。

旅行服务的表格完全不相关，并且不共享许多字段名称。

在合并搜索页面上，我正在尝试根据用户搜索字词从多个表中获取搜索结果。因此，如果用户输入“夏威夷”作为搜索词，他们应该会看到夏威夷作为停靠港，夏威夷的酒店，夏威夷的海滩度假村等游轮的结果。

例如，这里有3个表格（简化，现场网站上有许多其他字段）：

表格提出：

+----+--------------+-------------+-------+------+
| id | cruise_name  | destination | price | deck |
+----+--------------+-------------+-------+------+
|  1 | royal cruise | Hawaii      | 1000  | 1    |
|  2 | crown cruise | St. Martin  | 1200  | 2    |
+----+--------------+-------------+-------+------+

表酒店：

+----+--------------+-------------+-------+------+
| id | hotel_name   | location    | price | beds |
+----+--------------+-------------+-------+------+
|  1 | sheraton     | Hawaii      | 139   | 2    |
|  2 | meriott      | Florida     | 75    | 2    |
+----+--------------+-------------+-------+------+

表度假村：

+----+--------------+-------------+-------+-----------+
| id | resort       | location    | price | room_type |
+----+--------------+-------------+-------+-----------+
|  1 | sandals      | Jamaica     | 2000  | 3         |
|  2 | dream resort | Hawaii      | 2300  | 1         |
+----+--------------+-------------+-------+-----------+

作为一个简单的例子，假设我是用户搜索“夏威夷”，我需要向他们展示上面每个表格的结果。

我得到的最接近的是这样的查询：

SELECT  
    cuises.id AS cruise_id, 
    cuises.cruise_name AS cruise_cruise_name, 
    cuises.destination AS cruise_destination, 
    cuises.price AS cruise_price, 
    cuises.deck AS cruise_deck, 

    hotels.id AS hotels_id, 
    hotels.hotel_name AS hotels_hotel_name, 
    hotels.location AS hotels_location, 
    hotels.price AS hotels_price,
    hotels.beds AS hotels_beds,

    resorts.id AS resorts_id, 
    resorts.resort AS resorts_resort, 
    resorts.location AS resorts_location, 
    resorts.price AS resorts_price,
    resorts.room_type AS resorts_room_type

FROM 
    cuises,
    hotels,
    vacation_packages,
    resorts 

WHERE 
    (cuises.cruise_name LIKE '%Hawaii%') OR
    (hotels.hotel_name LIKE '%Hawaii%') OR
    (resorts.resort LIKE '%Hawaii%')

上面的样式查询将结果混合在一起，成为包含多个表中数据的行，而我正在寻找的是包含每行只有一个表格的信息的行。

我希望psuedo-tabel从查询中看起来像这样，这样我就可以轻松地在PHP中解析结果：

+-----------+---------------------+--------------------+--------------+-------------+-----------+---------------------+--------------------+--------------+-------------+------------+----------------------+---------------------+---------------+-------------------+
| cruise_id | cruise_cruise_name  | cruise_destination | cruise_price | cruise_deck | hotels_id | hotels_hotel_name   | hotels_location    | hotels_price | hotels_beds | resorts_id | resorts_resort       | resorts_location    | resorts_price | resorts_room_type |
+-----------+---------------------+--------------------+--------------+-------------+-----------+---------------------+--------------------+--------------+-------------+------------+----------------------+---------------------+---------------+-------------------+
|  1        | royal cruise        | Hawaii             | 1000         | 1           |           |                     |                    |              |             |            |                      |                     |               |                   |
|  2        | crown cruise        | St. Martin         | 1200         | 2           |           |                     |                    |              |             |            |                      |                     |               |                   |
|           |                     |                    |              |             |  1        | sheraton            | Hawaii             | 139          | 2           |            |                      |                     |               |                   |
|           |                     |                    |              |             |  2        | meriott             | Florida            | 75           | 2           |            |                      |                     |               |                   |
|           |                     |                    |              |             |           |                     |                    |              |             |  1         | sandals              | Jamaica             | 2000          | 3                 |
|           |                     |                    |              |             |           |                     |                    |              |             |  2         | dream resort         | Hawaii              | 2300          | 1                 |
+-----------+---------------------+--------------------+--------------+-------------+-----------+---------------------+--------------------+--------------+-------------+------------+----------------------+---------------------+---------------+-------------------+

基本上，如果我可以将搜索结果组合在多个表中，其中的字段就像上表一样放在一起，并且度假村的字段在有酒店信息的行中是空的，我可以轻松使用LIMIT和OFFSET对返回的数据进行分页，并使用PHP将返回的信息作为数组循环，并根据非空的id字段确定行中的旅行服务，以显示该行的信息使用适当的模板。

有没有办法查询多个表格，并将它们的行合并到一个不会混淆数据的结果集中？

Answer 1

您想要一个union query，这将在每个表上进行单独选择，并将所有结果合并到一个查询中。使用union查询的规定是所有union'd查询必须返回相同数量的列，因此您需要从每个子查询中的所有其他表select null as fieldname：

SELECT * FROM (
    (SELECT  
        cuises.id AS cruise_id, 
        cuises.cruise_name AS cruise_cruise_name, 
        cuises.destination AS cruise_destination, 
        cuises.price AS cruise_price, 
        cuises.deck AS cruise_deck,
        null as hotels_id, null as hotels_hotel_name, null as hotels_location, null as hotels_price, null as hotels_beds,
        null as resorts_id, null as resorts_resort, null as resorts_location, null as resorts_price, null as resorts_room_type
    FROM cruises
    WHERE cruises.cruise_name LIKE '%Hawaii%')

    UNION ALL

    (SELECT
        null as cruise_id, null as cruise_cruise_name, null as cruise_destination, null as cruise_price, null as cruise_deck,
        hotels.id AS hotels_id, 
        hotels.hotel_name AS hotels_hotel_name, 
        hotels.location AS hotels_location, 
        hotels.price AS hotels_price,
        hotels.beds AS hotels_beds
        null as resorts_id, null as resorts_resort, null as resorts_location, null as resorts_price, null as resorts_room_type
    FROM hotels
    WHERE hotels.hotel_name LIKE '%Hawaii%')

    UNION ALL

    (SELECT
        null as cruise_id, null as cruise_cruise_name, null as cruise_destination, null as cruise_price, null as cruise_deck,
        null as hotels_id, null as hotels_hotel_name, null as hotels_location, null as hotels_price, null as hotels_beds,
        resorts.id AS resorts_id, 
        resorts.resort AS resorts_resort, 
        resorts.location AS resorts_location, 
        resorts.price AS resorts_price,
        resorts.room_type AS resorts_room_type
    FROM resorts 
    WHERE resorts.resort LIKE '%Hawaii%')
)

从这里开始，您可以在此查询结束时轻松排序，过滤和限制结果。

注意：我没有在其中包含vacation_packages表，因为您没有从中选择任何内容或在条件中使用它。要添加它，只需使用新的UNION ALL语句添加另一个SELECT

Answer 2

您的查询正在执行各种属性的笛卡尔积。您只需要一个列表，因此union all是可行的方法。

问题是您必须为每个子选择定义所有列。呸。这需要输入显式的NULL值。以下显示了游轮和酒店的想法：

SELECT c.id AS cruise_id, c.cruise_name AS cruise_cruise_name, c.destination AS cruise_destination, 
       c.price AS cruise_price, c.deck AS cruise_deck, 
       NULL AS hotels_id, NULL AS hotels_hotel_name, NULL AS hotels_location, 
       NULL AS hotels_price, NULL AS hotels_beds,
       . . .
FROM cruises c
WHERE cuises.cruise_name LIKE '%Hawaii%'
union all
SELECT NULL AS cruise_id, NULLAS cruise_cruise_name, NULL AS cruise_destination, 
       NULL AS cruise_price, NULL AS cruise_deck, 
       h.id AS hotels_id,  h.hotel_name AS hotels_hotel_name, h.location AS hotels_location, 
       h.price AS hotels_price, h.beds AS hotels_beds,
       . . .
FROM hotels h
WHERE hotels.hotel_name LIKE '%Hawaii%';

Answer 3

我不认为你可以让mysql每个表返回1行，就像你想要的那样.. 我可能是错的，但我认为你真的不应该想要像这样返回你的数据。

我会使用LEFT JOIN和Alaises获取所有数据，并以您正在使用的任何显示语言解析它。

这是这些表的小提琴： http://sqlfiddle.com/#!2/7ebd7d/20

从包含不同和不相关信息的多个表中获取mysql搜索结果

3 个答案: