SQL東西錯過了..影響結果

Question

我已經在使用mysqli的這段代碼：

SELECT da_brands.name AS brand_name, 
    COUNT(da_deals.id) AS total_deals, 
    0 AS total_downloaded_coupons, 
    0 AS total_validated_coupons, 
    COUNT(da_logs.id) AS total_likes 
FROM da_brands, da_deals LEFT JOIN da_logs 
    ON da_logs.fk_deal_id = da_deals.id 
    AND da_logs.type = 'deal_like' 
WHERE da_brands.fk_club_id = 6 
    AND da_deals.fk_brand_id = da_brands.id 
GROUP BY da_brands.name 
ORDER BY da_brands.name ASC 

結果：

brand_name total_deals total_downloaded_coupons total_validated_coupons otal_likes 
Marca2  2    0       0       1 
Marca1  12    0       0       4 

應該是：

brand_name total_deals total_downloaded_coupons total_validated_coupons total_likes 
Marca2  2    0       0       1 
Marca1  9    0       0       4 

任何想法？

Answer 1

我相信你想要COUNT(DISTINCT da_deals.id) AS total_deals,，因爲你只想計算一次交易ID。

編輯：您的FROM語句是隱式內部連接（逗號）和顯式外部連接的列表。我錯過了。

您的查詢也許應該這樣寫，所有明確連接：

SELECT da_brands.name AS brand_name, 
    COUNT(DISTINCT da_deals.id) AS total_deals, 
    0 AS total_downloaded_coupons, 
    0 AS total_validated_coupons, 
    COUNT(DISTINCT da_logs.id) AS total_likes 
FROM da_brands 
INNER JOIN da_deals 
    ON da_deals.fk_brand_id = da_brands.id 
LEFT JOIN da_logs 
    ON da_logs.fk_deal_id = da_deals.id 
    AND da_logs.type = 'deal_like' 
WHERE da_brands.fk_club_id = 6 
GROUP BY da_brands.name 
ORDER BY da_brands.name ASC; 

爲了顯示品牌，即使他們缺乏交易和喜歡，嘗試LEFT JOIN這樣的：

SELECT da_brands.name AS brand_name, 
    COUNT(DISTINCT da_deals.id) AS total_deals, 
    0 AS total_downloaded_coupons, 
    0 AS total_validated_coupons, 
    COUNT(DISTINCT da_logs.id) AS total_likes 
FROM da_brands 
LEFT JOIN da_deals 
    ON da_deals.fk_brand_id = da_brands.id 
LEFT JOIN da_logs 
    ON da_logs.fk_deal_id = da_deals.id 
    AND da_logs.type = 'deal_like' 
WHERE da_brands.fk_club_id = 6 
GROUP BY da_brands.name 
ORDER BY da_brands.name ASC;