Question

我有3个表的问题：第一个（用户）包含汽车经销商，第二个和第三个（Order_new，Car_demo）包含经销商销售的两种类型的汽车。我需要总结经销商的销售数量和利润。不幸的是，它无法正常工作。其中一个经销商从Order_new出售了11辆汽车，从Car_demo出售了1辆，但是当我使用COUNT时，从Order_new（好）获得11，从Car_demo获得11（不好......）。与SUM相同 - 来自Car_demo我获得利润x 11。这是代码的重要片段：

SELECT
    d.name AS name,
    COUNT(cn_renault.id) AS ren_p,
    COUNT(cd_renault.id) AS ren_demo_p,
    SUM(cn_renault.profit) AS cn_profit,
    SUM(cd_renault.profit) AS cd_profit,
FROM User d
LEFT OUTER JOIN (
    SELECT DISTINCT
        cd2.id AS id,
        cd2.dealer AS sale_by,
        cd2.price AS price,
    FROM Order_new cd2
    LEFT OUTER JOIN Car car ON car.id = cd2.car_brand
    WHERE DATE(cd2.' . $by . ') >= \'' . $date . '\' AND DATE(cd2.' . $by . ') < \'' . $date_end . '\' AND car.brand = \'Renault\'
    GROUP BY cd2.id
) cn_renault ON cn_renault.sale_by = d.id
LEFT OUTER JOIN (
    SELECT DISTINCT
        cd2.id AS id,
        cd2.sale_by AS sale_by,
        cd2.selling_price AS price,
    FROM Car_demo cd2
    LEFT OUTER JOIN Car car ON car.id = cd2.car_brand
    WHERE DATE(cd2.' . $by_d . ') >= \'' . $date . '\' AND DATE(cd2.' . $by_d . ') < \'' . $date_end . '\' AND cd2.sale_status >= 2 AND car.brand = \'Renault\'
    GROUP BY cd2.id
) cd_renault ON cd_renault.sale_by = d.id
WHERE d.id = ' . $dealer_id . '
GROUP BY d.id

我知道我可以这样做：

LEFT OUTER JOIN (
        SELECT DISTINCT
            COUNT(cd2.id) AS number_of,
            SUM(cd2.price) AS price,
        FROM Order_new cd2
        LEFT OUTER JOIN Car car ON car.id = cd2.car_brand
        WHERE DATE(cd2.' . $by . ') >= \'' . $date . '\' AND DATE(cd2.' . $by . ') < \'' . $date_end . '\' AND car.brand = \'Renault\'
        GROUP BY cd2.dealer
    ) cn_renault ON cn_renault.sale_by = d.id

并且它有效，但我需要汽车ID才能在查询中加入另一个表...

有什么想法吗？

连接多个表时避免重复行

0 个答案: