需要帮助以优化 MySQL 查询答案

【问题标题】：Need Help for Optimizing MySQL query with few joins需要帮助以优化 MySQL 查询
【发布时间】：2015-02-24 13:51:46
【问题描述】：

您好，我有以下 sql 查询，它可以工作，但有点慢

SELECT 
    ep . *,
    t_u.sUsername AS sLockedBy,
    epi.sName,
    em.sName AS sManufacturerName,
    SUM(IF(eop.dInProduction IS NULL
            AND eop.dFromProduction IS NULL
            AND eop.dShipped IS NULL
            AND eop.nIsCancelled = 0
            AND eos.sStatusCode NOT IN ('Canceled' , 'Finished', 'Complete'),
        1,
        0)) AS nProductOrdered,
    SUM(IF(eop.dInProduction IS NOT NULL
            AND eop.dFromProduction IS NULL
            AND eop.dShipped IS NULL
            AND eop.nIsCancelled = 0
            AND eos.sStatusCode NOT IN ('Canceled' , 'Finished', 'Complete'),
        1,
        0)) AS nProductInProduction,
    SUM(IF(eop.dInProduction IS NOT NULL
            AND eop.dFromProduction IS NOT NULL
            AND eop.dShipped IS NULL
            AND eop.nIsCancelled = 0
            AND eos.sStatusCode NOT IN ('Canceled' , 'Finished', 'Complete'),
        1,
        0)) AS nProductFromProduction,
    SUM(IF(eop.dInProduction IS NULL
            AND eop.dFromProduction IS NULL
            AND eop.dShipped IS NULL
            AND eop.nIsCancelled = 0
            AND eos.sStatusCode NOT IN ('Canceled' , 'Finished', 'Complete'),
        1,
        0)) + ep.nInStock AS nStockCheck
FROM
    eshop3_products AS ep
        LEFT JOIN
    users AS t_u ON ep.nUserLockID = t_u.nUID
        AND t_u.nDeleted = 0
        INNER JOIN
    eshop3_products_i18n AS epi ON ep.nUID = epi.nProductID
        AND epi.nLangID = 1
        AND epi.nDeleted = 0
        LEFT JOIN
    eshop3_manufacturers AS em ON ep.nManufacturerID = em.nUID
        AND em.nDeleted = 0
        LEFT JOIN
    eshop3_order_products AS eop ON ep.nUID = eop.nProductID
        LEFT JOIN
    eshop3_orders AS eo ON eop.nOrderID = eo.nUID
        LEFT JOIN
    eshop3_order_statuses AS eos ON eo.nUID = eos.nOrderID
        AND eos.nUID = (SELECT 
            nUID
        FROM
            eshop3_order_statuses as eos
        WHERE
            eo.nUID = eos.nOrderID
        ORDER BY eos.nUID DESC
        LIMIT 1)
WHERE
    (ep.nDeleted = 0)
GROUP BY ep.nUID
ORDER BY ep.sStoreCode ASC

产品表共有 462 行，

订单状态表共有 17154 行

查询的执行时间是 1.5 秒，我觉得很慢。

我使用 ON 子句中的 INNER SELECT 来获取给定订单的最后状态，这里我认为有问题。

当我从整个查询中删除订单状态表时，执行时间快了好几倍，0.152 秒

你能给我一些建议，我该如何优化我的查询

这里是带有解释的 sql

id  select_type        table type   possible_keys    key      key_len ref                            rows  Extra
1   PRIMARY            epi   ALL    nProductID                                                        408  Using where; Using temporary; Using filesort
1   PRIMARY            ep    eq_ref PRIMARY,nUID     PRIMARY        4 dragoni_zfms.epi.nProductID       1  Using where
1   PRIMARY            em    eq_ref PRIMARY          PRIMARY        4 dragoni_zfms.ep.nManufacturerID   1 
1   PRIMARY            t_u   eq_ref PRIMARY          PRIMARY        4 dragoni_zfms.ep.nUserLockID       1 
1   PRIMARY            eop   ref    id_productid     id_productid   5 dragoni_zfms.ep.nUID             13 
1   PRIMARY            eo    eq_ref PRIMARY          PRIMARY        4 dragoni_zfms.eop.nOrderID         1  Using index
1   PRIMARY            eos   eq_ref PRIMARY,nOrderID PRIMARY        4 func                              1  
2   DEPENDENT SUBQUERY eos   ref    nOrderID         nOrderID       4 dragoni_zfms.eo.nUID              1  Using where; Using index; Using filesort

【问题讨论】：

尝试将该子选择更改为连接。也许与 group by 并选择 MAX 的东西会得到类似的结果，而不是 order by DESC LIMIT 1
您能否以文字而不是图片的形式提供说明。
请显示您的order_statuses 表的定义。
相同的别名 (eos) 用于标识不同的项目（实际上是同一张表，但在不同的上下文中被调用了两次）。也许这不是问题，但在调试时使用不同的名称会有所帮助。
@Strawberry 我已经提供了文字解释，你可以看看

标签： php mysql database database-optimization

【解决方案1】：

在您的情况下，不使用 mysql 而是使用 php 执行一些操作会更容易。当你赢得时间时，我在我的练习案例中看到过。请尝试将一个复杂的查询分成几个简单的查询。然后使用 php 脚本执行必要的过滤。希望能帮助到你。谢谢！

【讨论】：