Question

我是stackoverflow中的新手。我的查询花了很多时间（超过5分钟）。这个查询可以变得更快吗？请！帮我！我不知道:(

我有2个表插座和销售： 1.出口：具有唯一数据的表格 - id_outlet - name_outlet 2.销售：像插座的细节（1个插座有很多数据） - id_outlet - msisdn（1个出口超过1 msisdn） - 日期

所以，我需要获取数据： 1.分支 2.插座数量 3.活动插座数量（我从本月检查现有插座获得此数据） 3.计数msisdn 4.后退检查插座的数量（我从上个月检查现有插座获得此数据，但本月不存在） 5.新活动插座的数量（我从本月检查现有插座获得此数据，但在上个月不存在） 6.非活动插座的数量（我从上个月和本月检查该插座不存在获得此数据） 7. Consistant Outlet的数量（我从本月和上个月检查现有商店获得此数据）

这是我的查询：

SELECT branch br, COUNT(DISTINCT id_outlet) AS tot_outlet,
                    (SELECT COUNT(DISTINCT(id_outlet))
                        FROM outlet o
                        WHERE ".$outlet['status'][$i]." AND active=1 AND branch = br
                        AND id_outlet IN (SELECT DISTINCT id_outlet FROM sales WHERE DATE LIKE '".$thisMonth."%')) AS aktif,
                    (SELECT COUNT(msisdn) FROM sales s, outlet o
                        WHERE ".$outlet['status'][$i]." and s.date LIKE '".$thisMonth."%' AND o.active=1 AND s.id_outlet = o.id_outlet
                         AND s.branch=br) AS supply,
                    (SELECT COUNT(DISTINCT(id_outlet))
                        FROM outlet o
                        WHERE ".$outlet['status'][$i]." AND active=1 AND branch = br
                        AND id_outlet IN (SELECT DISTINCT id_outlet FROM sales WHERE DATE LIKE '".$lastMonth."%')
                        AND id_outlet NOT IN (SELECT DISTINCT id_outlet FROM sales WHERE DATE LIKE '".$thisMonth."%')) AS back_checking,
                    (SELECT COUNT(DISTINCT(id_outlet))
                        FROM outlet o
                        WHERE ".$outlet['status'][$i]." AND active=1 AND branch = br
                        AND id_outlet NOT IN (SELECT DISTINCT id_outlet FROM sales WHERE DATE LIKE '".$lastMonth."%')
                        AND id_outlet IN (SELECT DISTINCT id_outlet FROM sales WHERE DATE LIKE '".$thisMonth."%')) AS new_aktif,
                    (SELECT COUNT(DISTINCT(id_outlet))
                        FROM outlet o
                        WHERE ".$outlet['status'][$i]." AND active=1 AND branch = br
                        AND id_outlet NOT IN (SELECT DISTINCT id_outlet FROM sales WHERE DATE LIKE '".$lastMonth."%')
                        AND id_outlet NOT IN (SELECT DISTINCT id_outlet FROM sales WHERE DATE LIKE '".$thisMonth."%')) AS non_aktif,
                    (SELECT COUNT(DISTINCT(id_outlet))
                        FROM outlet o
                        WHERE ".$outlet['status'][$i]." AND active=1 AND branch = br
                        AND id_outlet IN (SELECT DISTINCT id_outlet FROM sales WHERE DATE LIKE '".$lastMonth."%')
                        AND id_outlet IN (SELECT DISTINCT id_outlet FROM sales WHERE DATE LIKE '".$thisMonth."%')) AS konsisten
                FROM outlet o
                WHERE ".$outlet['status'][$i]." AND active=1 
                GROUP BY branch

Answer 1

自己诊断的几个步骤。

解释您的疑问。

要解释您的查询，这很简单;您只需要在EXPLAIN部分的前面添加SELECT。

运行此;

EXPLAIN SELECT branch br, COUNT(DISTINCT id_outlet) AS tot_outlet,
                (SELECT COUNT(DISTINCT(id_outlet))
                    FROM outlet o
                    WHERE ".$outlet['status'][$i]." AND active=1 AND branch = br
                    AND id_outlet IN (SELECT DISTINCT id_outlet FROM sales WHERE DATE LIKE '".$thisMonth."%')) AS aktif,
                (SELECT SUM(featurephone) FROM outlet o
                    WHERE ".$outlet['status'][$i]." 
                    AND active=1 AND branch = br) AS feature,
                (SELECT SUM(smartphone) FROM outlet o
                    WHERE ".$outlet['status'][$i]." 
                    AND active=1 AND branch = br) AS smart,
                (SELECT COUNT(msisdn) FROM sales s, outlet o
                    WHERE ".$outlet['status'][$i]." and s.date LIKE '".$thisMonth."%' AND o.active=1 AND s.id_outlet = o.id_outlet
                     AND s.branch=br) AS supply,
                (SELECT COUNT(DISTINCT(id_outlet))
                    FROM outlet o
                    WHERE ".$outlet['status'][$i]." AND active=1 AND branch = br
                    AND id_outlet IN (SELECT DISTINCT id_outlet FROM sales WHERE DATE LIKE '".$lastMonth."%')
                    AND id_outlet NOT IN (SELECT DISTINCT id_outlet FROM sales WHERE DATE LIKE '".$thisMonth."%')) AS back_checking,
                (SELECT COUNT(DISTINCT(id_outlet))
                    FROM outlet o
                    WHERE ".$outlet['status'][$i]." AND active=1 AND branch = br
                    AND id_outlet NOT IN (SELECT DISTINCT id_outlet FROM sales WHERE DATE LIKE '".$lastMonth."%')
                    AND id_outlet IN (SELECT DISTINCT id_outlet FROM sales WHERE DATE LIKE '".$thisMonth."%')) AS new_aktif,
                (SELECT COUNT(DISTINCT(id_outlet))
                    FROM outlet o
                    WHERE ".$outlet['status'][$i]." AND active=1 AND branch = br
                    AND id_outlet NOT IN (SELECT DISTINCT id_outlet FROM sales WHERE DATE LIKE '".$lastMonth."%')
                    AND id_outlet NOT IN (SELECT DISTINCT id_outlet FROM sales WHERE DATE LIKE '".$thisMonth."%')) AS non_aktif,
                (SELECT COUNT(DISTINCT(id_outlet))
                    FROM outlet o
                    WHERE ".$outlet['status'][$i]." AND active=1 AND branch = br
                    AND id_outlet IN (SELECT DISTINCT id_outlet FROM sales WHERE DATE LIKE '".$lastMonth."%')
                    AND id_outlet IN (SELECT DISTINCT id_outlet FROM sales WHERE DATE LIKE '".$thisMonth."%')) AS konsisten
            FROM outlet o
            WHERE ".$outlet['status'][$i]." AND active=1 
            GROUP BY branch

这将提供类似于此的输出：

Explain View

简要说明您应该寻找什么。最重要的是type，possible_keys和key。主要是type。

Type的值可以告诉您它是如何查找查询的。列表（按从最好到最差的顺序）是：system, const, eq_ref, ref, fulltext, ref_or_null, unique_subquery, index_subquery, range, index, all。你不想要任何全部;而且最有可能的是，主要关注ref和eq_ref。由于查询的结构，您可能会有子查询。

要解决这个问题，我要做的是查看表的结构和查询。任何将从排序列表（基本上是索引）中受益的东西都会被索引。对我来说，这意味着我将索引任何外键ID（即使你不使用外键，设置索引）;任何属性，如删除，可见或其他。

简单地说，我将大多数在查询的where部分中进行比较的内容编入索引。这是一个好的开始;从那里，你真的应该优化的东西，因为索引可以使用大量的内存，但这是性能提升的良好开端。

Answer 2

避免在子查询中使用子查询，那么你应该能够做到这样的事情： -

SELECT branch br, COUNT(DISTINCT id_outlet) AS tot_outlet,
                    (SELECT COUNT(DISTINCT(id_outlet))
                        FROM outlet o
                        INNER JOIN sales s2 ON o.id_outlet = s2.id_outlet AND s2.DATE LIKE '".$thisMonth."%'
                        WHERE ".$outlet['status'][$i]." 
                        AND o.active=1 
                        AND o.branch = br) AS aktif,
                    (SELECT COUNT(msisdn) FROM sales s, outlet o
                        WHERE ".$outlet['status'][$i]." 
                        and s.date LIKE '".$thisMonth."%' 
                        AND o.active=1 
                        AND s.id_outlet = o.id_outlet
                         AND s.branch=br) AS supply,
                    (SELECT COUNT(DISTINCT(id_outlet))
                        FROM outlet o
                        INNER JOIN sales s1 ON o.id_outlet = s1.id_outlet AND s1.DATE LIKE '".$lastMonth."%'
                        LEFT OUTER JOIN sales s2 ON o.id_outlet = s2.id_outlet AND s2.DATE LIKE '".$thisMonth."%'
                        WHERE ".$outlet['status'][$i]." 
                        AND o.active=1 
                        AND o.branch = br
                        AND s2.id_outlet IS NULL) AS back_checking,
                    (SELECT COUNT(DISTINCT(id_outlet))
                        FROM outlet o
                        LEFT OUTER JOIN sales s1 ON o.id_outlet = s1.id_outlet AND s1.DATE LIKE '".$lastMonth."%'
                        INNER JOIN sales s2 ON o.id_outlet = s2.id_outlet AND s2.DATE LIKE '".$thisMonth."%'
                        WHERE ".$outlet['status'][$i]." 
                        AND o.active=1 
                        AND o.branch = br
                        AND s1.id_outlet IS NULL) AS new_aktif,
                    (SELECT COUNT(DISTINCT(o.id_outlet))
                        FROM outlet o
                        LEFT OUTER JOIN sales s1 ON o.id_outlet = s1.id_outlet AND s1.DATE LIKE '".$lastMonth."%'
                        LEFT OUTER JOIN sales s2 ON o.id_outlet = s2.id_outlet AND s2.DATE LIKE '".$thisMonth."%'
                        WHERE ".$outlet['status'][$i]." 
                        AND o.active=1 
                        AND o.branch = br
                        AND s1.id_outlet IS NULL
                        AND s2.id_outlet IS NULL) AS non_aktif,
                    (SELECT COUNT(DISTINCT(o.id_outlet))
                        FROM outlet o
                        INNER JOIN sales s1 ON o.id_outlet = s1.id_outlet AND s1.DATE LIKE '".$lastMonth."%'
                        INNER JOIN sales s2 ON o.id_outlet = s2.id_outlet AND s2.DATE LIKE '".$thisMonth."%'
                        WHERE ".$outlet['status'][$i]." AND o.active=1 AND o.branch = br) AS konsisten
                FROM outlet o
                WHERE ".$outlet['status'][$i]." AND active=1 
                GROUP BY branch

这仍然在SELECT中使用多个子查询，但是根据WHERE子句（即$ outlet ['status'] [$ i]的值），这可能不会太糟糕。

我知道你说你已经做到了这一点，但它给出了错误的结果，但是如果没有示例数据和表格布局，我无法做任何检查以确定这应该是错误的原因或原因。

有可能将连接放在主查询中，至少替换最后4个子查询： -

SELECT branch br, COUNT(DISTINCT id_outlet) AS tot_outlet,
                    (SELECT COUNT(DISTINCT(id_outlet))
                        FROM outlet o
                        INNER JOIN sales s2 ON o.id_outlet = s2.id_outlet AND s2.DATE LIKE '".$thisMonth."%'
                        WHERE ".$outlet['status'][$i]." 
                        AND o.active=1 
                        AND o.branch = br) AS aktif,
                    (SELECT COUNT(msisdn) FROM sales s, outlet o
                        WHERE ".$outlet['status'][$i]." 
                        and s.date LIKE '".$thisMonth."%' 
                        AND o.active=1 
                        AND s.id_outlet = o.id_outlet
                         AND s.branch=br) AS supply,
                    COUNT(DISTINCT IF(s1.id_outlet IS NOT NULL AND s2.id_outlet IS NULL, o.id_outlet, NULL) AS back_checking
                    COUNT(DISTINCT IF(s1.id_outlet IS NULL AND s2.id_outlet IS NOT NULL, o.id_outlet, NULL) AS new_aktif
                    COUNT(DISTINCT IF(s1.id_outlet IS NULL AND s2.id_outlet IS NULL, o.id_outlet, NULL) AS non_aktif
                    COUNT(DISTINCT IF(s1.id_outlet IS NOT NULL AND s2.id_outlet IS NOT NULL, o.id_outlet, NULL) AS konsisten
                FROM outlet o
                LEFT OUTER JOIN sales s1 ON o.id_outlet = s1.id_outlet AND s1.DATE LIKE '".$lastMonth."%'
                LEFT OUTER JOIN sales s2 ON o.id_outlet = s2.id_outlet AND s2.DATE LIKE '".$thisMonth."%'
                WHERE ".$outlet['status'][$i]." AND active=1 
                GROUP BY branch

这依赖于仅计算非空值的COUNT（某个表达式）。

EDIT。

一个选项是将核心子查询更改为非核心子查询，您可以对其进行连接。然后，每次只执行一次，而不是每次返回一次。缺点是加入子查询结果不会使用索引，因此连接可能不会那么快。哪个更有效取决于数据量。

使用我之前的查询并更改atkif和supply字段以使用非核心子查询，可以得到： -

SELECT o.branch br, 
    COUNT(DISTINCT o.id_outlet) AS tot_outlet, 
    COALESCE(sub_aktif.aktif, 0) AS aktif, 
    COALESCE(sub_supply.supply, 0) AS supply, 
    COUNT(DISTINCT IF(s1.id_outlet IS NOT NULL AND s2.id_outlet IS NULL, o.id_outlet, NULL)) AS back_checking,
    COUNT(DISTINCT IF(s1.id_outlet IS NULL AND s2.id_outlet IS NOT NULL, o.id_outlet, NULL)) AS new_aktif,
    COUNT(DISTINCT IF(s1.id_outlet IS NULL AND s2.id_outlet IS NULL, o.id_outlet, NULL)) AS non_aktif,
    COUNT(DISTINCT IF(s1.id_outlet IS NOT NULL AND s2.id_outlet IS NOT NULL, o.id_outlet, NULL)) AS konsisten 
FROM outlet o 
LEFT OUTER JOIN
(
    SELECT o.branch, COUNT(DISTINCT(o.id_outlet)) AS aktif
    FROM outlet o 
    INNER JOIN sales s 
    ON o.id_outlet = s.id_outlet 
    WHERE (o.STATUS LIKE 'MALL%' OR o.STATUS LIKE 'STREET%' OR o.STATUS LIKE 'TOP TEN%' OR o.STATUS LIKE 'LIFESTYLE%') 
    AND s.DATE LIKE '2014-10%' AND o.active=1 
    GROUP BY o.branch 
) sub_aktif
ON sub_aktif.branch = o.branch
LEFT OUTER JOIN
(
    SELECT s.branch, COUNT(msisdn) AS supply
    FROM sales s
    INNER JOIN outlet o 
    ON s.id_outlet = o.id_outlet
    WHERE (o.STATUS LIKE 'MALL%' OR o.STATUS LIKE 'STREET%' OR o.STATUS LIKE 'TOP TEN%' OR o.STATUS LIKE 'LIFESTYLE%') 
    AND s.date LIKE '2014-10%' AND o.active=1 
    GROUP BY s.branch 
) sub_supply
ON sub_supply.branch = o.branch
LEFT OUTER JOIN sales s1 ON o.id_outlet = s1.id_outlet AND s1.DATE LIKE '2014-09%' 
LEFT OUTER JOIN sales s2 ON o.id_outlet = s2.id_outlet AND s2.DATE LIKE '2014-10%' 
WHERE (o.STATUS LIKE 'MALL%' OR o.STATUS LIKE 'STREET%' OR o.STATUS LIKE 'TOP TEN%' OR o.STATUS LIKE 'LIFESTYLE%') 
AND active=1 
GROUP BY o.branch ;

更改原始查询以对所有字段执行此操作会产生以下结果： -

SELECT o.branch br, COUNT(DISTINCT id_outlet) AS tot_outlet,
    COALESCE(sub_aktif.aktif, 0) AS aktif, 
    COALESCE(sub_supply.supply, 0) AS supply, 
    COALESCE(sub_back_checking.back_checking, 0) AS back_checking, 
    COALESCE(sub_new_aktif.new_aktif, 0) AS new_aktif, 
    COALESCE(sub_non_aktif.non_aktif, 0) AS non_aktif, 
    COALESCE(sub_konsisten.konsisten, 0) AS konsisten
FROM outlet o
LEFT OUTER JOIN
(
    SELECT o.branch, COUNT(DISTINCT(o.id_outlet)) AS aktif
    FROM outlet o 
    INNER JOIN sales s 
    ON o.id_outlet = s.id_outlet 
    WHERE (o.STATUS LIKE 'MALL%' OR o.STATUS LIKE 'STREET%' OR o.STATUS LIKE 'TOP TEN%' OR o.STATUS LIKE 'LIFESTYLE%') 
    AND s.DATE LIKE '2014-10%' AND o.active=1 
    GROUP BY o.branch 
) sub_aktif
ON sub_aktif.branch = o.branch
LEFT OUTER JOIN
(
    SELECT s.branch, COUNT(msisdn) AS supply
    FROM sales s
    INNER JOIN outlet o 
    ON s.id_outlet = o.id_outlet
    WHERE (o.STATUS LIKE 'MALL%' OR o.STATUS LIKE 'STREET%' OR o.STATUS LIKE 'TOP TEN%' OR o.STATUS LIKE 'LIFESTYLE%') 
    AND s.date LIKE '2014-10%' AND o.active=1 
    GROUP BY s.branch 
) sub_supply
ON sub_supply.branch = o.branch
LEFT OUTER JOIN
(
    SELECT o.branch, COUNT(DISTINCT(o.id_outlet)) AS back_checking
    FROM outlet o
    INNER JOIN sales s1
    ON s1.id_outlet = o.id_outlet AND s1.DATE LIKE '2014-09%'
    LEFT OUTER JOIN sales s2
    ON s2.id_outlet = o.id_outlet AND s2.DATE LIKE '2014-10%'
    WHERE (o.STATUS LIKE 'MALL%' OR o.STATUS LIKE 'STREET%' OR o.STATUS LIKE 'TOP TEN%' OR o.STATUS LIKE 'LIFESTYLE%') AND active=1 
    AND s2.id_outlet IS NULL
    GROUP BY o.branch
) sub_back_checking
ON sub_back_checking.branch = o.branch
LEFT OUTER JOIN
(
    SELECT o.branch, COUNT(DISTINCT(o.id_outlet)) AS new_aktif
    FROM outlet o
    LEFT OUTER JOIN sales s1
    ON s1.id_outlet = o.id_outlet AND s1.DATE LIKE '2014-09%'
    INNER JOIN sales s2
    ON s2.id_outlet = o.id_outlet AND s2.DATE LIKE '2014-10%'
    WHERE (o.STATUS LIKE 'MALL%' OR o.STATUS LIKE 'STREET%' OR o.STATUS LIKE 'TOP TEN%' OR o.STATUS LIKE 'LIFESTYLE%') AND active=1 
    AND s1.id_outlet IS NULL
    GROUP BY o.branch
) sub_new_aktif
ON sub_new_aktif.branch = o.branch
LEFT OUTER JOIN
(
    SELECT o.branch, COUNT(DISTINCT(o.id_outlet)) AS non_aktif
    FROM outlet o
    LEFT OUTER JOIN sales s1
    ON s1.id_outlet = o.id_outlet AND s1.DATE LIKE '2014-09%'
    LEFT OUTER JOIN sales s2
    ON s2.id_outlet = o.id_outlet AND s2.DATE LIKE '2014-10%'
    WHERE (o.STATUS LIKE 'MALL%' OR o.STATUS LIKE 'STREET%' OR o.STATUS LIKE 'TOP TEN%' OR o.STATUS LIKE 'LIFESTYLE%') AND active=1 
    AND s1.id_outlet IS NULL
    AND s2.id_outlet IS NULL
    GROUP BY o.branch
) sub_non_aktif
ON sub_non_aktif.branch = o.branch
LEFT OUTER JOIN
(
    SELECT o.branch, COUNT(DISTINCT(o.id_outlet)) AS konsisten
    FROM outlet o
    INNER JOIN sales s1
    ON s1.id_outlet = o.id_outlet AND s1.DATE LIKE '2014-09%'
    INNER JOIN sales s2
    ON s2.id_outlet = o.id_outlet AND s2.DATE LIKE '2014-10%'
    WHERE (o.STATUS LIKE 'MALL%' OR o.STATUS LIKE 'STREET%' OR o.STATUS LIKE 'TOP TEN%' OR o.STATUS LIKE 'LIFESTYLE%') AND active=1 
    GROUP BY o.branch
) sub_konsisten
ON sub_konsisten.branch = o.branch
WHERE (o.STATUS LIKE 'MALL%' OR o.STATUS LIKE 'STREET%' OR o.STATUS LIKE 'TOP TEN%' OR o.STATUS LIKE 'LIFESTYLE%') AND active=1 
GROUP BY o.branch;

sql小提琴在这里： -

http://sqlfiddle.com/#!2/1343e/20

MySQL查询时间过长，花费了大量时间

这是我的查询：

2 个答案: