（SQL）匹配用戶屬於哪個組給定的user_id []

用戶表（SQL）匹配用戶屬於哪個組給定的user_id []

ID | name 
1 | ada  
2 | bob 
3 | tom

組表

ID | name 
1 | group A 
2 | group B 
3 | group C

USER_GROUP表

user_id | group_id 
1  | 1 
2  | 1 
1  | 2 
2  | 2 
3  | 2 
1  | 3 
3  | 3

給定組的用戶ID： 1，2，3]

t o查詢上述列表中的所有用戶屬於哪個組？（在這種情況下：B組）

來源

2013-07-22 David Ng

如果你能保證'USER_ID，group_id'是獨一無二的 - 那麼'COUNT' +'GROUP BY '+'HAVING' – zerkms

user_id 2如何屬於組C.我認爲組b是你所需要的？ –

是，B組，感謝Dinup –

要獲得正好包含指定用戶的所有組（即所有指定的用戶和沒有其他用戶）

DECLARE @numUsers int = 3 

SELECT ug.group_id 
     --The Max doesn't really do anything here because all 
     --groups with the same group id have the same name. The 
     --max is just used so we can select the group name eventhough 
     --we aren't aggregating across group names 
    , MAX(g.name) AS name 
FROM user_group ug 
--Filter to only groups with three users 
JOIN (SELECT group_id FROM user_group GROUP BY group_id HAVING COUNT(*) = @numUsers) ug2 
    ON ug.group_id = ug2.group_id 
JOIN [group] g 
    ON ug.group_id = g.ID 
WHERE user_id IN (1, 2, 3) 
GROUP BY ug.group_id 
--The distinct is only necessary if user_group 
--isn't keyed by group_id, user_id 
HAVING COUNT(DISTINCT user_id) = @numUsers

要獲得包含所有指定的用戶組：

DECLARE @numUsers int = 3  

    SELECT ug.group_id 
      --The Max doesn't really do anything here because all 
      --groups with the same group id have the same name. The 
      --max is just used so we can select the group name eventhough 
      --we aren't aggregating across group names 
     , MAX(g.name) AS name 
    FROM user_group ug 
    JOIN [group] g 
     ON ug.group_id = g.ID 
    WHERE user_id IN (1, 2, 3) 
    GROUP BY ug.group_id 
    --The distinct is only necessary if user_group 
    --isn't keyed by group_id, user_id 
    HAVING COUNT(DISTINCT user_id) = 3

SQL小提琴：http://sqlfiddle.com/#!6/0e968/3

來源

2013-07-22 05:12:43

我認爲你不需要加入 – FUD

但是這隻有在zerkms提出的是正確的時纔有效。 – FUD

@FUD'JOIN'只是爲了從OPT表中獲取信息而獲取信息。 –

Select UserID,count(*) 
From UserGroupTable 
group by UserID

這將給3在用戶ID /羣ID是唯一的計數（如zerkms指出）

來源

2013-07-22 05:09:24 user1516509

這不是一個goodsolution如果我只想要[1,2]您的解決方案將返回第1組和第3，這是錯的 – FUD

問題詢問「給user_id []」..所以WHERE子句/或類似的限制查詢的方式是必要的。 –

SELECT name FROM group_tbl WHERE id IN (SELECT g_id FROM user_grp GROUP BY g_id HAVING Count(u_id)=(SELECT Count(id) FROM user_tbl));

來源

2013-07-22 05:48:27 Aspirant

試試這個：

Select t2.name   
    FROM   
    (Select group_id 
     From 
     user_group 
     Group by group_id 
    Having Count(user_id) = (Select Count(*) FROM User_Table)) AS T1   
    INNER JOIN   
     Group_Table AS T2 
     ON T1.group_id = T2.ID

見小提琴：http://sqlfiddle.com/#!2/fa7250/4

來源

2013-07-22 06:11:55 Transformer

這不是一個通用的解決方案，如果我們想知道組用戶[1,3]都在... –

我還以爲你要檢查所有用戶表中可用的mebers。 – Transformer

對不起，我做到了不清楚 –

（SQL）匹配用戶屬於哪個組給定的user_id []

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