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我有很多表，我从中选择，但我只会写重要部分。我有 2 张桌子

住宿：

id | title | desc | etc..

住宿房间：

id | accomodation_id | beds | rooms

然后我一直选择哪个效果好，直到我想过滤，有多少床（床 * 房间）有住宿。如果我尝试使用乘法从表 accomodation_rooms 中进行简单选择，则效果很好。所以问题应该出在我加入其他表的地方。

这是我的选择：

SELECT 
  accomodation.*, 
  db_cities.title_en AS city, 
  db_cities.title_url AS city_url, 
  db_countries.title_url_en AS country_url, 
  SUM(accomodation_rooms.beds * accomodation_rooms.rooms) AS total_persons 
FROM 
  (SELECT id, aid, title_en, title_url_en, address, city_id, zip, district_id, province_id, region_id, country_id, mountain_id, stars, latitude, longitude, picture, valid_from, valid_to 
  FROM accomodation 
  ORDER BY info_date_add DESC) 
    AS accomodation 
LEFT JOIN db_cities 
  ON db_cities.id = accomodation.city_id 
JOIN db_countries 
  ON db_countries.id = accomodation.country_id 
JOIN skiresort_locations 
  ON 
    (((acos(sin((skiresort_locations.latitude*pi()/180)) * 
    sin((accomodation.latitude*pi()/180)) + 
    cos((skiresort_locations.latitude*pi()/180)) * 
    cos((accomodation.latitude*pi()/180)) * 
    cos(((skiresort_locations.longitude - 
    accomodation.longitude)*
    pi()/180))))*180/pi())*60*1.1515*1.609344) 
      < '50' 
JOIN accomodation_rooms 
  ON accomodation_rooms.accomodation_id = accomodation.id 
WHERE 
  db_countries.title_url_en LIKE '%country_title%' AND 
  accomodation.region_id = '8' 
GROUP BY 
  accomodation.aid 
HAVING 
  total_persons >= '1' 
ORDER BY CASE 
  WHEN 
    accomodation.valid_to>=NOW() AND 
    accomodation.valid_from<=NOW() 
      THEN 0 
  WHEN 
    NOW()>accomodation.valid_to AND 
    accomodation.valid_to!='0000-00-00' 
      THEN 1 
  ELSE 2 
    END, 
  accomodation.title_en 
LIMIT 10 
OFFSET 0

total_persons 应该返回 13 但它返回 624，但我真的不明白为什么？

4

2 回答 2

1

认为您想要的数量可能最好通过使用子选择来获得。

我假设 accomodation_rooms 是一个表，其中存储了一些住宿的多行，说明该位置有多少张床（即 1 x 5 床房间、2 x 3 床、5 x 2 床）。

SELECT 
  accomodation.*, 
  db_cities.title_en AS city, 
  db_cities.title_url AS city_url, 
  db_countries.title_url_en AS country_url, 
  accomodation_rooms.total_persons 
FROM 
  (SELECT id, aid, title_en, title_url_en, address, city_id, zip, district_id, province_id, region_id, country_id, mountain_id, stars, latitude, longitude, picture, valid_from, valid_to 
  FROM accomodation 
  ORDER BY info_date_add DESC) 
    AS accomodation 
LEFT JOIN db_cities 
  ON db_cities.id = accomodation.city_id 
JOIN db_countries 
  ON db_countries.id = accomodation.country_id 
JOIN skiresort_locations 
  ON 
    (((acos(sin((skiresort_locations.latitude*pi()/180)) * 
    sin((accomodation.latitude*pi()/180)) + 
    cos((skiresort_locations.latitude*pi()/180)) * 
    cos((accomodation.latitude*pi()/180)) * 
    cos(((skiresort_locations.longitude - 
    accomodation.longitude)*
    pi()/180))))*180/pi())*60*1.1515*1.609344) 
      < '50' 
INNER JOIN (SELECT accomodation_id, SUM(accomodation_rooms.beds * accomodation_rooms.rooms) AS total_persons FROM accomodation_rooms GROUP BY accomodation_id) accomodation_rooms
  ON accomodation_rooms.accomodation_id = accomodation.id 
  AND accomodation_rooms.total_persons >= 1
WHERE db_countries.title_url_en LIKE '%country_title%' 
AND accomodation.region_id = '8' 
ORDER BY CASE 
  WHEN accomodation.valid_to>=NOW() AND accomodation.valid_from<=NOW() 
      THEN 0 
  WHEN NOW()>accomodation.valid_to AND accomodation.valid_to!='0000-00-00' 
      THEN 1 
  ELSE 2 
    END, 
  accomodation.title_en 
LIMIT 10 
OFFSET 0

作为一个小问题，不确定您是否需要将原始选择作为子选择（或有一个 order by 子句，但您似乎没有限制，所以我认为它不会有任何影响） :-

SELECT 
  accomodation.id, 
  accomodation.aid, 
  accomodation.title_en, 
  accomodation.title_url_en, 
  accomodation.address, 
  accomodation.city_id, 
  accomodation.zip, 
  accomodation.district_id, 
  accomodation.province_id, 
  accomodation.region_id, 
  accomodation.country_id, 
  accomodation.mountain_id, 
  accomodation.stars, 
  accomodation.latitude, 
  accomodation.longitude, 
  accomodation.picture, 
  accomodation.valid_from, 
  accomodation.valid_to, 
  db_cities.title_en AS city, 
  db_cities.title_url AS city_url, 
  db_countries.title_url_en AS country_url, 
  accomodation_rooms.total_persons 
FROM accomodation
LEFT JOIN db_cities 
  ON db_cities.id = accomodation.city_id 
JOIN db_countries 
  ON db_countries.id = accomodation.country_id 
JOIN skiresort_locations 
  ON 
    (((acos(sin((skiresort_locations.latitude*pi()/180)) * 
    sin((accomodation.latitude*pi()/180)) + 
    cos((skiresort_locations.latitude*pi()/180)) * 
    cos((accomodation.latitude*pi()/180)) * 
    cos(((skiresort_locations.longitude - 
    accomodation.longitude)*
    pi()/180))))*180/pi())*60*1.1515*1.609344) 
      < '50' 
INNER JOIN (SELECT accomodation_id, SUM(accomodation_rooms.beds * accomodation_rooms.rooms) AS total_persons FROM accomodation_rooms GROUP BY accomodation_id) accomodation_rooms
  ON accomodation_rooms.accomodation_id = accomodation.id 
  AND accomodation_rooms.total_persons >= 1
WHERE db_countries.title_url_en LIKE '%country_title%' 
AND accomodation.region_id = '8' 
ORDER BY CASE 
  WHEN accomodation.valid_to>=NOW() AND accomodation.valid_from<=NOW() 
      THEN 0 
  WHEN NOW()>accomodation.valid_to AND accomodation.valid_to!='0000-00-00' 
      THEN 1 
  ELSE 2 
    END, 
  accomodation.title_en 
LIMIT 10 OFFSET 0

编辑 - 修改以获取每个援助的最新住宿记录，并加入该记录以获取最新住宿记录的其余部分。

SELECT 
  accomodation.*, 
  db_cities.title_en AS city, 
  db_cities.title_url AS city_url, 
  db_countries.title_url_en AS country_url, 
  accomodation_rooms.total_persons 
FROM (SELECT aid, MAX(info_date_add) AS max_info_date_add FROM accomodation GROUP BY aid) accomodation_max
INNER JOIN accomodation ON accomodation_max.aid = accomodation.aid AND accomodation_max.max_info_date_add = accomodation.info_date_add
LEFT JOIN db_cities ON db_cities.id = accomodation.city_id 
JOIN db_countries ON db_countries.id = accomodation.country_id 
JOIN skiresort_locations 
  ON 
    (((acos(sin((skiresort_locations.latitude*pi()/180)) * 
    sin((accomodation.latitude*pi()/180)) + 
    cos((skiresort_locations.latitude*pi()/180)) * 
    cos((accomodation.latitude*pi()/180)) * 
    cos(((skiresort_locations.longitude - 
    accomodation.longitude)*
    pi()/180))))*180/pi())*60*1.1515*1.609344) 
      < '50' 
INNER JOIN (SELECT accomodation_id, SUM(accomodation_rooms.beds * accomodation_rooms.rooms) AS total_persons FROM accomodation_rooms GROUP BY accomodation_id) accomodation_rooms
ON accomodation_rooms.accomodation_id = accomodation.id AND accomodation_rooms.total_persons >= 1
WHERE db_countries.title_url_en LIKE '%country_title%' 
AND accomodation.region_id = '8' 
ORDER BY CASE 
  WHEN accomodation.valid_to>=NOW() AND accomodation.valid_from<=NOW() 
      THEN 0 
  WHEN NOW()>accomodation.valid_to AND accomodation.valid_to!='0000-00-00' 
      THEN 1 
  ELSE 2 
    END, 
  accomodation.title_en 
LIMIT 10 
OFFSET 0

编辑 - 使用 GROUP BY 子句为距离添加 MIN。但是不确定这会比使用 DISTINCT 快得多。它会强制对 JOIN 进行大量计算（即，如果您有 100 条住宿记录和 100 条skiresort_locations 记录，那么这将导致 10000 次合理的复杂计算来确定距离。如果您可以在计算，那么这将节省相当多的时间（例如，去规范化一点，但也许你有一个区域表连接到 region_id 上的住宿，它可能包含该区域的最小和最大纬度和经度，你可以使用对于连接，然后将复杂的计算放入 WHERE 子句中）。还有一个用于 db_counties 的 LIKE 子句。

SELECT 
  accomodation.*, 
  db_cities.title_en AS city, 
  db_cities.title_url AS city_url, 
  db_countries.title_url_en AS country_url, 
  accomodation_rooms.total_persons,
  MIN(((acos(sin((skiresort_locations.latitude*pi()/180)) * 
    sin((accomodation.latitude*pi()/180)) + 
    cos((skiresort_locations.latitude*pi()/180)) * 
    cos((accomodation.latitude*pi()/180)) * 
    cos(((skiresort_locations.longitude - 
    accomodation.longitude)*
    pi()/180))))*180/pi())*60*1.1515*1.609344)
FROM (SELECT aid, MAX(info_date_add) AS max_info_date_add FROM accomodation GROUP BY aid) accomodation_max
INNER JOIN accomodation ON accomodation_max.aid = accomodation.aid AND accomodation_max.max_info_date_add = accomodation.info_date_add
LEFT JOIN db_cities ON db_cities.id = accomodation.city_id 
JOIN db_countries ON db_countries.id = accomodation.country_id 
JOIN skiresort_locations 
  ON 
    (((acos(sin((skiresort_locations.latitude*pi()/180)) * 
    sin((accomodation.latitude*pi()/180)) + 
    cos((skiresort_locations.latitude*pi()/180)) * 
    cos((accomodation.latitude*pi()/180)) * 
    cos(((skiresort_locations.longitude - 
    accomodation.longitude)*
    pi()/180))))*180/pi())*60*1.1515*1.609344) 
      < '50' 
INNER JOIN (SELECT accomodation_id, SUM(accomodation_rooms.beds * accomodation_rooms.rooms) AS total_persons FROM accomodation_rooms GROUP BY accomodation_id) accomodation_rooms
ON accomodation_rooms.accomodation_id = accomodation.id AND accomodation_rooms.total_persons >= 1
WHERE db_countries.title_url_en LIKE '%country_title%' 
AND accomodation.region_id = '8' 
GROUP BY accomodation.id, 
  db_cities.title_en, 
  db_cities.title_url, 
  db_countries.title_url, 
  accomodation_rooms.total_persons
ORDER BY CASE 
  WHEN accomodation.valid_to>=NOW() AND accomodation.valid_from<=NOW() 
      THEN 0 
  WHEN NOW()>accomodation.valid_to AND accomodation.valid_to!='0000-00-00' 
      THEN 1 
  ELSE 2 
    END, 
  accomodation.title_en 
LIMIT 10 
OFFSET 0
于 2013-06-20T13:27:28.997 回答
1

对新要求的进一步答复。

您需要从找到的住宿重新加入到城市表中以获取该住宿的城市详细信息：-

SELECT 
   accomodation.id, 
   accomodation.aid,
   accomodation.title_$lang, 
   accomodation.title_url_$lang,
   accomodation.address,
   accomodation.zip, 
   accomodation.stars, 
   accomodation.picture, 
   accomodation.valid_from, 
   accomodation.valid_to, 
   de_cities_accomodation.title_$lang AS city,
   de_cities_accomodation.title_url AS city_url, 
   db_countries.title_url_$lang AS country_url
FROM  db_cities 
INNER JOIN accomodation 
   ON 
   ( 
      db_cities.id = accomodation.city_id 
      OR 
      (((acos(sin((db_cities.latitude*pi()/180)) * sin((accomodation.latitude*pi()/180)) + cos((db_cities.latitude*pi()/180)) * cos((accomodation.latitude*pi()/180)) * cos(((db_cities.longitude - accomodation.longitude)*pi()/180))))*180/pi())*60*1.1515*1.609344) < '20' ) 
INNER JOIN (SELECT aid, MAX(info_date_add) AS max_info_date_add FROM accomodation GROUP BY aid) accomodation_max 
   ON 
      accomodation_max.aid = accomodation.aid AND 
      accomodation_max.max_info_date_add = accomodation.info_date_add 
LEFT JOIN db_cities de_cities_accomodation
   ON de_cities_accomodation.id = accomodation.city_id 
JOIN db_countries 
   ON db_countries.id = accomodation.country_id 
WHERE 
   db_countries.title_url_en LIKE '%country%' AND 
   db_cities.title_url LIKE '%city%' AND 
   db_cities.id = '1258' 
GROUP BY accomodation.aid 
ORDER BY 
   CASE 
      WHEN accomodation.valid_to>=NOW() AND accomodation.valid_from<=NOW() THEN 0 
     WHEN NOW()>accomodation.valid_to AND accomodation.valid_to!='0000-00-00' THEN 1 
   ELSE 2 END
于 2013-07-01T15:29:54.133 回答