SQL GROUP BY重複行

Question

我有一個Postges數據庫一個很長的桌子和3列，像這樣：

s_id | c_id | a_id 
1 | 1 | 2 
1 | 1 | 3 
1 | 3 | 15 
2 | 1 | 2 
2 | 2 | 23 
3 | 1 | 2 
3 | 3 | 16 

我有一個查詢發現已C_ID 1和3的所有s_ids，返回他們和他們的計數：

SELECT s_id, COUNT(s_id) as matching_clusters 
FROM test 
WHERE c_id IN (1,3) 
GROUP BY s_id HAVING COUNT(c_id) >= 2 
ORDER BY matching_clusters DESC 

我得到的回覆是以下幾點：

s_id | matching_clusters 
1 |   3 
3 |   2 

但是，我只想計算重複C_ID一次，這樣的結果在這裏應該是

s_id | matching_clusters 
1 |   2 
3 |   2 

任何建議如何做到這一點？我認爲我可以將DISTINCT放入COUNT命令中，但這不起作用。我可以用不同的c_id在表本身上加入結果，但我不想重新運行查詢，因爲在這個表上運行查詢是非常昂貴的計算明智的。

Answer 1

如果我理解正確的話，那麼這將工作：

SELECT s_id, 2 as matching_clusters 
FROM test 
WHERE c_id IN (1,3) 
GROUP BY s_id 
HAVING COUNT(c_id) >= 2 
ORDER BY matching_clusters DESC; 

這可能是你想要什麼：

SELECT s_id, COUNT(DISTINCT c_id) as matching_clusters 
FROM test 
WHERE c_id IN (1,3) 
GROUP BY s_id 
HAVING COUNT(DISTINCT c_id) = 2 
ORDER BY matching_clusters DESC; 

注意having子句中使用distinct。

Answer 2

試試這個： -

SELECT s_id, COUNT(DISTINCT s_id) as matching_clusters 
FROM test 
WHERE c_id IN (1,3) 
GROUP BY s_id HAVING COUNT(c_id) >= 2 
ORDER BY matching_clusters DESC