Muliple基於兩種不同的評論加入總值

Question

我有一個有4個表的數據庫。

Table 1 - "company" table with company_id as the key 
Table 2 - "users" table with the user_id as the key 
Table 3 - "teams" table that references the company_id and the user_id (So that user can belong to multiple teams. 
Table 4 - "points" table that references the company_id, the user_id, points_earned (Numeric value of points given), exchange_dte (0 - if the user has not used the points, otherwise a unixtime value) 

給定一個已知的company_id，我想打電話給屬於該「團隊」的所有用戶，並顯示其總分和其未交換點。以下MySQL只會給公司的＃1團隊中的第一個用戶。目前數據庫中有5個用戶都獲得了點數，有些是交換的，有些則沒有。

SELECT 
users.user_id AS u_id, 
SUM(points.points_earned) AS ttl_points, 
SUM(case when exchange_dte = '0' then points.points_earned else 0 end) AS unused_points 
FROM users 
INNER JOIN teams ON teams.user_id = users.user_id 
INNER JOIN points ON points.user_id = users.user_id 
WHERE (teams.company_id = '1' AND points.company_id = '1' AND users.user_active = '1'); 

所以，然後我試着添加user_id到總和調用。並以同樣的事情結束。

SELECT 
users.user_id AS u_id, 
SUM(case when points.user_id = users.user_id then points.points_earned else 0 end) AS ttl_points, 
SUM(case when points.exchange_dte = '0' AND points.user_id = users.user_id then points.points_earned else 0 end) AS unused_points 
FROM users 
INNER JOIN teams ON teams.user_id = users.user_id 
INNER JOIN points ON points.user_id = users.user_id 
WHERE (teams.company_id = '1' AND points.company_id = '1' AND users.user_active = '1') 
ORDER BY ttl_points; 

有趣的是，第一個用戶點總數似乎是數據庫中的所有的點，即使他們有他們

思考相關的USER_ID和COMPANY_ID？

Answer 1

你正在試圖做一個沒有使用GROUP BY的SUM：不知道它是否適用於你，但試着在查詢結束後添加GROUP BY users.user_id，看看是否能幫上你。

SELECT 
users.user_id AS u_id, 
SUM(case when points.user_id = users.user_id then points.points_earned else 0 end) AS ttl_points, 
SUM(case when points.exchange_dte = '0' AND points.user_id = users.user_id then points.points_earned else 0 end) AS unused_points 
FROM users 
INNER JOIN teams ON teams.user_id = users.user_id 
INNER JOIN points ON points.user_id = users.user_id 
WHERE (teams.company_id = '1' AND points.company_id = '1' AND users.user_active = '1') GROUP BY users.user_id 
ORDER BY ttl_points;