SQL查詢從朋友

發現簽入我有一個PostgreSQL下表：SQL查詢從朋友

Categories | Locations | Checkins  | Users | Friendships 
id   | id   | id   | id  | user_id 
name  | category_id | location_id | gender | friend_id 
icon  | name  | user_id  |  |

現在，我想檢索有關場地的下列信息

多少女性和男性用戶一個位置都有
類別名稱和圖標
地點名稱
有幾個朋友CHEC ked在一個位置（從給定的用戶ID）

除了最後一點，我解決了它。但我有麻煩來計算來自給定用戶ID的朋友。我試着用這個查詢：

SELECT distinct locations.id, 
max(locations.name) as name, 
max(locations.location) as location, 
max(categories.name) as cat, 
max(categories.icon) as caticon, 
SUM(CASE WHEN users.gender = 'm' THEN 1 ELSE 0 END) AS male, 
SUM(CASE WHEN users.gender = 'f' THEN 1 ELSE 0 END) AS female, 
SUM(CASE WHEN friendships.user_id = 1 OR friendships.friend_id=1 THEN 1 ELSE 0 END) AS friends 
FROM locations 
INNER JOIN checkins ON checkins.location_id = locations.id 
INNER JOIN users ON users.id = checkins.user_id 
INNER JOIN categories ON categories.id = locations.category_id 
LEFT JOIN friendships ON friendships.user_id = users.id OR friendships.friend_id = users.id 
WHERE locations.id=7 
GROUP BY locations.id

但我得到一個錯誤的女性用戶數。任何想法我做錯了什麼？我想我需要一個友誼表的左連接，因爲如果用戶沒有朋友（或沒有用戶給出），它應該只返回0爲好友數。

希望我自己清楚， THX，禮服

來源

2011-05-22 23tux

'從地點'，我相信它應該是。此外，'WHITE friendships.user_id = 1 OR friendships.user_id = 1'應該可以是'WHITE friendships.user_id = 1 OR friendships.friend_id = 1'。我有一個問題：如果兩個用戶是彼此的朋友，那麼在'友誼'中會有多少記錄，1或2？並且2條記錄應該產生1或2個友誼？ – 2011-05-22 10:28:37

thx爲您的答案（我固定它，它是早上toooo;））。當兩個用戶是彼此的朋友時，友誼表中只有一條記錄，請求友誼的用戶存儲在'user_id'中，另一個用於'friend_id' – 23tux 2011-05-22 10:31:01

SELECT 
    L.id, 
    L.name, 
    c.name AS cat, 
    c.icon AS caticon, 
    COUNT(CASE u.gender WHEN 'm' THEN 1 END) AS male, 
    COUNT(CASE u.gender WHEN 'f' THEN 1 END) AS female, 
    COUNT(f.user_id) AS friends 
FROM Locations L 
    INNER JOIN Categories c ON c.id = L.category_id 
    INNER JOIN Checkins ch ON ch.location_id = L.id 
    INNER JOIN Users u ON u.id = ch.user_id 
    LEFT JOIN Friendships f ON f.user_id = @user_id AND f.friend_id = ch.user_id 
          OR f.user_id = ch.user_id AND f.friend_id = @user_id 
WHERE L.id = @location_id 
GROUP BY L.id, L.name, c.name, c.icon

來源

2011-05-22 10:50:56

真棒，解決了！ ;） – 23tux 2011-05-23 09:15:54

降distinct在第一線。對於同一字段，您已經有一個group by子句。讓我知道這是否有幫助。

來源

2011-05-22 10:28:42 Jai

SQL查詢從朋友

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