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这是我的查询(已编辑)
SELECT
en.name AS name,
en.entity_id,
COUNT(o.order_id) AS orders_qty,
ROUND(SUM(o.total)) AS orders_sum,
ROUND((SUM(o.total) / COUNT(o.order_id))) AS average_purchase,
MIN(o.date_added) AS first_purchase,
MAX(o.date_added) AS last_purchase,
(SELECT COUNT(*) FROM oc_order WHERE order_status_id <> 0 AND customer_id = c.customer_id AND date_added <= NOW() - INTERVAL 3 MONTH) as periodicity
FROM oc_xile_entity en
LEFT JOIN oc_xile_customer_to_entity c2en ON (c2en.entity_id = en.entity_id)
LEFT JOIN oc_customer c ON (c2en.customer_id = c.customer_id)
LEFT JOIN oc_order o FORCE INDEX FOR JOIN (`unreg_customer_id`) ON ((o.customer_id = c2en.customer_id OR o.unreg_customer_id = c2en.customer_id) AND order_status_id <> 0)
WHERE en.entity_id IS NOT NULL
GROUP BY en.entity_id
ORDER BY name ASC
LIMIT 0,700
...以及表格的相关部分...
CREATE TABLE oc_order
( order_id int(11) NOT NULL PRIMARY KEY
, customer_id int(11) NOT NULL DEFAULT '0'
, unreg_customer_id (unreg_customer_id)
, unreg_customer_id int(11) NOT NULL
, order_status_id int(11) NOT NULL DEFAULT '0'
, INDEX (customer_id)
);
CREATE TABLE oc_customer
(customer_id int(11) NOT NULL PRIMARY KEY);
CREATE TABLE oc_xile_customer_to_entity
(entity_id int(11) NOT NULL
,customer_id int(11) NOT NULL
,PRIMARY KEY (entity_id,customer_id)
,INDEX entity_id
);
当我查询表顺序 35506 行时,它花费了超过 15 秒
最繁重的查询是
LEFT JOIN oc_order o ON ((o.customer_id = c2en.customer_id OR o.unreg_customer_id = c2en.customer_id) AND order_status_id 0)
最好的优化方式是添加FORCE INDEX FOR JOIN (unreg_customer_id)
像这样LEFT JOIN oc_order o FORCE INDEX FOR JOIN (unreg_customer_id) ON ((o.customer_id = c2en.customer_id OR o.unreg_customer_id = c2en.customer_id) AND order_status_id <> 0)
其速度提升,查询时间变为3秒。但我认为它可以优化得更好。有谁能够帮助我?查询的解释
最佳答案
呃?该查询对 LEFT JOIN 不执行任何操作。然而,优化器可能太笨了,无法丢弃它们并获得等效的查询:
SELECT en.name AS name , en.entity_id
FROM oc_xile_entity en
WHERE en.entity_id IS NOT NULL
GROUP BY en.entity_id
ORDER BY name ASC
LIMIT 0 , 700
那么这可能是“不正确的”,因为您选择了名称
,但没有按它进行分组。如果有多个名称
与一个entity_id
关联,那么应该显示哪个名称
?
请描述您想要实现的目标,并提供SHOW CREATE TABLE oc_xile_entity
。
关于mysql - 通过 OR 优化左连接?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/54442074/
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