Php 从被推荐的用户那里获取支付金额的总和

我正在开发一个推荐系统，作为该仪表板的一部分，我需要输出用户收到的推荐数量、他们因这些推荐购买物品而获得的硬币总数，以及最终购买物品的推荐数量和总额

例如，显示用户_ID 1的信息：

我的表格结构如下：

使用者

购买

目前我正在运行这个查询。出于某种原因，我无法理解这个问题，在本例中，它只获取与id=2匹配的第一个用户的和。countid=3，但sumamount=10.00，而不是预期的14.00。有什么想法吗

SELECT count(id) AS count, 
(SELECT COALESCE(sum(amount), 0.00) 
    FROM purchases
    WHERE purchases.user_id = users.id 
    AND payment_status = "Completed" 
) AS purchases_amount, 
(SELECT COALESCE(sum(round(amount) * 10), 0) 
    FROM purchases
    WHERE users.id = purchases.user_id 
    AND payment_status = "Completed"
) AS coins_earned,    
(SELECT count(purchase_id) 
    FROM purchases
    WHERE users.id = purchases.user_id
    AND payment_status = "Completed"
) AS purchases_count
FROM (`users`) 
WHERE `referred_by` = '1' 
AND `created` >= '2015-01-01 00:00:00' // created = user's registration date
AND `created` <= '2015-12-31 23:59:59'

---

您遇到了MySQL问题，它允许聚合与非聚合在没有适当分组的情况下进行组合。其他RDBMS会抱怨您的查询，因为第一列中的计数聚合与查询其余部分返回的多行冲突

尝试删除select countid AS count中的第一列，您将看到查询为每个用户id 2、3和4返回一行

要解决此问题，请将每个子查询的所有结果相加：

SELECT count(id) AS count, 
sum((SELECT COALESCE(sum(amount), 0.00) 
    FROM purchases
    WHERE purchases.user_id = users.id 
    AND payment_status = "Completed" 
)) AS purchases_amount, 
sum((SELECT COALESCE(sum(round(amount) * 10), 0) 
    FROM purchases
    WHERE users.id = purchases.user_id 
    AND payment_status = "Completed"
)) AS coins_earned,    
sum((SELECT count(purchase_id) 
    FROM purchases
    WHERE users.id = purchases.user_id
    AND payment_status = "Completed"
)) AS purchases_count
FROM (`users`) 
WHERE `referred_by` = '1' 
AND `created` >= '2015-01-01 00:00:00' // created = user's registration date
AND `created` <= '2015-12-31 23:59:59'

---

我相信你可以试试这个

SELECT 
    COALESCE(SUM(amount), 0.00) AS purchases_amount,
    COALESCE(SUM(ROUND(amount) * 10), 0) AS coins_earned,
    COUNT(purchase_id) AS purchases_count
FROM purchases
WHERE
    payment_status = "Completed"     
    AND user_id IN (
        SELECT id FROM users
        WHERE `referred_by` = '1' 
        AND `created` >= '2015-01-01 00:00:00'
        AND `created` <= '2015-12-31 23:59:59'  
    )

---

这似乎可以从一些SQL简化和分组中获益

SELECT count(id) AS count, 
(SELECT COALESCE(sum(amount), 0.00) 
    FROM purchases
    WHERE purchases.user_id = users.id 
    AND payment_status = "Completed" 
) AS purchases_amount, 
(SELECT COALESCE(sum(round(amount) * 10), 0) 
    FROM purchases
    WHERE users.id = purchases.user_id 
    AND payment_status = "Completed"
) AS coins_earned,    
(SELECT count(purchase_id) 
    FROM purchases
    WHERE users.id = purchases.user_id
    AND payment_status = "Completed"
) AS purchases_count
FROM (`users`) 
WHERE `referred_by` = '1' 
AND `created` >= '2015-01-01 00:00:00' // created = user's registration date
AND `created` <= '2015-12-31 23:59:59'

count: 3 (expected: 3)
purchases_amount: 10.00 (expected: 14.00)
coins_earned: 100 (expected: 140)
purchases_count: 1 (expected: 2)

SELECT count(id) AS count, 
sum((SELECT COALESCE(sum(amount), 0.00) 
    FROM purchases
    WHERE purchases.user_id = users.id 
    AND payment_status = "Completed" 
)) AS purchases_amount, 
sum((SELECT COALESCE(sum(round(amount) * 10), 0) 
    FROM purchases
    WHERE users.id = purchases.user_id 
    AND payment_status = "Completed"
)) AS coins_earned,    
sum((SELECT count(purchase_id) 
    FROM purchases
    WHERE users.id = purchases.user_id
    AND payment_status = "Completed"
)) AS purchases_count
FROM (`users`) 
WHERE `referred_by` = '1' 
AND `created` >= '2015-01-01 00:00:00' // created = user's registration date
AND `created` <= '2015-12-31 23:59:59'

SELECT 
    COALESCE(SUM(amount), 0.00) AS purchases_amount,
    COALESCE(SUM(ROUND(amount) * 10), 0) AS coins_earned,
    COUNT(purchase_id) AS purchases_count
FROM purchases
WHERE
    payment_status = "Completed"     
    AND user_id IN (
        SELECT id FROM users
        WHERE `referred_by` = '1' 
        AND `created` >= '2015-01-01 00:00:00'
        AND `created` <= '2015-12-31 23:59:59'  
    )