MySQL count rows where another column = value

Question

我有一個While循環，在每行中顯示COUNT的p.songid（後軌）。我實際需要的是根據trackDeleted=0而不是所有行來計算計數。

SELECT u.username, u.id, u.score, s.genre, s.songid, s.songTitle, s.timeSubmitted, s.userid, s.insWanted, s.bounty, COUNT(p.songid) 
FROM songs s 
LEFT JOIN users u ON u.id = s.userid 
LEFT JOIN posttracks p ON s.songid = p.songid 
WHERE paid=1 AND s.timeSubmitted >= (CURDATE() - INTERVAL 60 DAY) 
GROUP BY s.timeSubmitted ASC 
LIMIT 25 

我可以更新上述查詢，還是必須在單獨的查詢中？

Answer 1

它取決於trackDeleted列在哪個表中。

如果它生活在posttracks，那麼它的簡單

SELECT COUNT(p.songid) FROM posttracks p where p.trackDeleted=0 

Answer 2

添加trackDeleted = 0您WHERE條款：

... 
WHERE paid=1 AND s.timeSubmitted >= (CURDATE() - INTERVAL 60 DAY) 
    AND trackDeleted=0 
-- ^^^ add this 
GROUP BY s.timeSubmitted ASC 
... 

Answer 3

嘗試...... 您必須由p.songid組。

SELECT u.username, u.id, u.score, s.genre, s.songid, s.songTitle, 
s.timeSubmitted, s.userid, s.insWanted, s.bounty, COUNT(p.songid) FROM 
songs s LEFT JOIN users u ON u.id = s.userid LEFT JOIN posttracks p ON 
s.songid = p.songid WHERE paid=1 AND s.timeSubmitted >= (CURDATE() - 
INTERVAL 60 DAY) GROUP BY s.timeSubmitted ASC,p.songid LIMIT 25 

Answer 4

試試這個。

DECLARE @trackCount int, 
SELECT u.username, u.id, u.score, s.genre, s.songid, s.songTitle, s.timeSubmitted, s.userid, s.insWanted, s.bounty, 
FROM songs s 
LEFT JOIN users u ON u.id = s.userid 
LEFT JOIN posttracks p ON s.songid = p.songid 
WHERE paid=1 
AND s.timeSubmitted >= (CURDATE() - INTERVAL 60 DAY) 
GROUP BY s.timeSubmitted ASC 
LIMIT 25 

SELECT COUNT(p.songid) INTO @TrackCount 
FROM posttracks p 
WHERE p.trackDeleted = 0 

嘗試將此代碼添加到您的代碼中。只需訪問本地變量@trackcount，如果您需要顯示有多少曲目trackDeleted = 0。

Answer 5

如果您要COUNT歌曲ID，您必須通過它自己對它進行分組，嘗試這一個。

SELECT u.username, u.id, u.score, s.genre, s.songid, s.songTitle, s.timeSubmitted, s.userid, s.insWanted, s.bounty, COUNT(p.songid) 
    FROM songs s 
    LEFT JOIN users u 
    ON u.id = s.userid 
    LEFT JOIN posttracks p 
    ON s.songid = p.songid 
    WHERE paid=1 
    AND s.timeSubmitted >= (CURDATE() - INTERVAL 60 DAY) 
    AND trackDeleted=0 
    GROUP BY p.songid 
    ORDER BY s.timeSubmitted 
    LIMIT 25; 

Answer 6

要麼把你的測試中加入條件：

LEFT JOIN posttracks p 
    ON s.songid = p.songid 
    AND trackDeleted = 0 

或更改COUNT(p.songid)的款項，而不是：

SUM(CASE WHEN (p.songid IS NOT NULL AND trackDeleted = 0) THEN 1 ELSE 0 END) 

Answer 7

萬一有人絆倒這一點，像我一樣，正確答案可能類似於：

SELECT u.username, u.id, u.score, s.genre, s.songid, s.songTitle, s.timeSubmitted, s.userid, s.insWanted, s.bounty, COUNT(IF(p.trackDeleted=0, 1, NULL)) as not_deleted_count 
    FROM songs s 
    LEFT JOIN users u 
    ON u.id = s.userid 
    LEFT JOIN posttracks p 
    ON s.songid = p.songid 
    WHERE paid=1 AND s.timeSubmitted >= (CURDATE() - INTERVAL 60 DAY) 
    GROUP BY s.timeSubmitted ASC 
    LIMIT 25 

關鍵是COUNT(IF(p.trackDeleted=0, 1, NULL)) as not_deleted_count它將查看分組中的所有曲目，並只計算那些未被刪除的曲目。

在這裏的每一個其他答案都誤認爲OP只想選擇未刪除的曲目，而當他想要的是統計在每個分組內部刪除的曲目數量。

Answer 8

我不知道，如果這個工程在MySQL，但在SQL你能過濾過程中加入如下：

SELECT u.username, u.id, u.score, s.genre, s.songid, s.songTitle, s.timeSubmitted, s.userid, s.insWanted, s.bounty, COUNT(p.songid) 
FROM songs s 
LEFT JOIN users u ON u.id = s.userid 
LEFT JOIN posttracks p ON s.songid = p.songid and trackDeleted=0 
WHERE paid=1 AND s.timeSubmitted >= (CURDATE() - INTERVAL 60 DAY) 
GROUP BY s.timeSubmitted ASC 
LIMIT 25