我的例子是 MYSQL VERSION是 34年5月6日日志
问题摘要以下查询需要 40秒， ORDER_ITEM 表
有 758423 条记录
付款表
有 177272 记录
submission_entry 表
有 2165698 条记录
作为整个表计数。
DETAILS HERE：BELOW：
我有此查询，请参阅 [1]
我已添加 SQL_NO_CACHE ，以便在重新测试时测试重复测试 查询。
我有优化的索引参考 [2] ，但没有重要意义 改善。
在此处查找表格结构 [3]
[1]
SELECT SQL_NO_CACHE
`payment`.`id` AS id,
`order_item`.`order_id` AS order_id,
GROUP_CONCAT(DISTINCT (CASE WHEN submission_entry.text = '' OR submission_entry.text IS NULL
THEN ' '
ELSE submission_entry.text END) ORDER BY question.var DESC SEPARATOR 0x1D) AS buyer,
event.name AS event,
COUNT(DISTINCT CASE WHEN (`order_item`.status > 0 OR (
`order_item`.status != -1 AND `order_item`.status >= -2 AND `payment`.payment_type_id != 8 AND
payment.make_order_free = 1))
THEN `order_item`.id
ELSE NULL END) AS qty,
payment.currency AS `currency`,
(SELECT SUM(order_item.sub_total)
FROM order_item
WHERE payment_id =
payment.id) AS sub_total,
CASE WHEN payment.make_order_free = 1
THEN ROUND(payment.total + COALESCE(refunds_total, 0), 2)
ELSE ROUND(payment.total, 2) END AS 'total',
`payment_type`.`name` AS payment_type,
payment_status.name AS status,
`payment_status`.`id` AS status_id,
DATE_FORMAT(CONVERT_TZ(order_item.`created`, '+0:00', '-8:00'),
'%Y-%m-%d %H:%i') AS 'created',
`user`.`name` AS 'agent',
event.id AS event_id,
payment.checked,
DATE_FORMAT(CONVERT_TZ(payment.checked_date, '+0:00', '-8:00'),
'%Y-%m-%d %H:%i') AS checked_date,
DATE_FORMAT(CONVERT_TZ(`payment`.`complete_date`, '+0:00', '-8:00'),
'%Y-%m-%d %H:%i') AS `complete date`,
`payment`.`delivery_status` AS `delivered`
FROM `order_item`
INNER JOIN `payment`
ON payment.id = `order_item`.`payment_id` AND (payment.status > 0.0 OR payment.status = -3.0)
LEFT JOIN (SELECT
sum(`payment_refund`.total) AS `refunds_total`,
payment_refunds.payment_id AS `payment_id`
FROM payment
INNER JOIN `payment_refunds` ON payment_refunds.payment_id = payment.id
INNER JOIN `payment` AS `payment_refund`
ON `payment_refund`.id = `payment_refunds`.payment_id_refund
GROUP BY `payment_refunds`.payment_id) AS `refunds` ON `refunds`.payment_id = payment.id
# INNER JOIN event_date_product ON event_date_product.id = order_item.event_date_product_id
# INNER JOIN event_date ON event_date.id = event_date_product.event_date_id
INNER JOIN event ON event.id = order_item.event_id
INNER JOIN payment_status ON payment_status.id = payment.status
INNER JOIN payment_type ON payment_type.id = payment.payment_type_id
LEFT JOIN user ON user.id = payment.completed_by
LEFT JOIN submission_entry ON submission_entry.form_submission_id = `payment`.`form_submission_id`
LEFT JOIN question ON question.id = submission_entry.question_id AND question.var IN ('name', 'email')
WHERE 1 = '1' AND (order_item.status > 0.0 OR order_item.status = -2.0)
GROUP BY `order_item`.`order_id`
HAVING 1 = '1'
ORDER BY `order_item`.`order_id` DESC
LIMIT 10
[2]
CREATE INDEX order_id
ON order_item (order_id);
CREATE INDEX payment_id
ON order_item (payment_id);
CREATE INDEX status
ON order_item (status);
第二张表
CREATE INDEX payment_type_id
ON payment (payment_type_id);
CREATE INDEX status
ON payment (status);
[3]
CREATE TABLE order_item
(
id INT AUTO_INCREMENT
PRIMARY KEY,
order_id INT NOT NULL,
form_submission_id INT NULL,
status DOUBLE DEFAULT '0' NULL,
payment_id INT DEFAULT '0' NULL
);
SECOND TABLE
CREATE TABLE payment
(
id INT AUTO_INCREMENT,
payment_type_id INT NOT NULL,
status DOUBLE NOT NULL,
form_submission_id INT NOT NULL,
PRIMARY KEY (id, payment_type_id)
);
[4] 运行代码段以HTML格式查看 EXPLAIN 表格
<!DOCTYPE html>
<html>
<head>
<title></title>
</head>
<body>
<table border="1" style="border-collapse:collapse">
<tr><th>id</th><th>select_type</th><th>table</th><th>type</th><th>possible_keys</th><th>key</th><th>key_len</th><th>ref</th><th>rows</th><th>Extra</th></tr>
<tr><td>1</td><td>PRIMARY</td><td>payment_status</td><td>range</td><td>PRIMARY</td><td>PRIMARY</td><td>8</td><td>NULL</td><td>4</td><td>Using where; Using temporary; Using filesort</td></tr>
<tr><td>1</td><td>PRIMARY</td><td>payment</td><td>ref</td><td>PRIMARY,payment_type_id,status</td><td>status</td><td>8</td><td>exp_live_18092017.payment_status.id</td><td>17357</td><td></td></tr>
<tr><td>1</td><td>PRIMARY</td><td>payment_type</td><td>eq_ref</td><td>PRIMARY</td><td>PRIMARY</td><td>4</td><td>exp_live_18092017.payment.payment_type_id</td><td>1</td><td></td></tr>
<tr><td>1</td><td>PRIMARY</td><td>user</td><td>eq_ref</td><td>PRIMARY</td><td>PRIMARY</td><td>4</td><td>exp_live_18092017.payment.completed_by</td><td>1</td><td></td></tr>
<tr><td>1</td><td>PRIMARY</td><td>submission_entry</td><td>ref</td><td>form_submission_id,idx_submission_entry_1</td><td>form_submission_id</td><td>4</td><td>exp_live_18092017.payment.form_submission_id</td><td>2</td><td></td></tr>
<tr><td>1</td><td>PRIMARY</td><td>question</td><td>eq_ref</td><td>PRIMARY,var</td><td>PRIMARY</td><td>4</td><td>exp_live_18092017.submission_entry.question_id</td><td>1</td><td>Using where</td></tr>
<tr><td>1</td><td>PRIMARY</td><td>order_item</td><td>ref</td><td>status,payment_id</td><td>payment_id</td><td>5</td><td>exp_live_18092017.payment.id</td><td>3</td><td>Using where</td></tr>
<tr><td>1</td><td>PRIMARY</td><td>event</td><td>eq_ref</td><td>PRIMARY</td><td>PRIMARY</td><td>4</td><td>exp_live_18092017.order_item.event_id</td><td>1</td><td></td></tr>
<tr><td>1</td><td>PRIMARY</td><td><derived3></td><td>ref</td><td>key0</td><td>key0</td><td>5</td><td>exp_live_18092017.payment.id</td><td>10</td><td>Using where</td></tr>
<tr><td>3</td><td>DERIVED</td><td>payment_refunds</td><td>index</td><td>payment_id,payment_id_refund</td><td>payment_id</td><td>4</td><td>NULL</td><td>1110</td><td></td></tr>
<tr><td>3</td><td>DERIVED</td><td>payment</td><td>ref</td><td>PRIMARY</td><td>PRIMARY</td><td>4</td><td>exp_live_18092017.payment_refunds.payment_id</td><td>1</td><td>Using index</td></tr>
<tr><td>3</td><td>DERIVED</td><td>payment_refund</td><td>ref</td><td>PRIMARY</td><td>PRIMARY</td><td>4</td><td>exp_live_18092017.payment_refunds.payment_id_refund</td><td>1</td><td></td></tr>
<tr><td>2</td><td>DEPENDENT SUBQUERY</td><td>order_item</td><td>ref</td><td>payment_id</td><td>payment_id</td><td>5</td><td>func</td><td>3</td><td></td></tr></table>
</body>
</html>
预期的Restul
必须代替40秒，而不是5
重要 的更新
1）回复评论1：这两个表上根本没有外键。
UPDATE-1： 在本地上，原始查询需要 40秒 如果我删除 以下内容 25秒保存 15秒
GROUP_CONCAT(DISTINCT (CASE WHEN submission_entry.text = '' OR submission_entry.text IS NULL
THEN ' '
ELSE submission_entry.text END) ORDER BY question.var DESC SEPARATOR 0x1D) AS buyer
如果我在 40秒
如果我删除 ，则需要 36秒保存 4秒 COUNT(DISTINCT CASE WHEN (`order_item`.status > 0 OR (
`order_item`.status != -1 AND `order_item`.status >= -2 AND `payment`.payment_type_id != 8 AND
payment.make_order_free = 1))
THEN `order_item`.id
ELSE NULL END) AS qty,
(SELECT SUM(order_item.sub_total)
FROM order_item
WHERE payment_id =
payment.id) AS sub_total,
CASE WHEN payment.make_order_free = 1
THEN ROUND(payment.total + COALESCE(refunds_total, 0), 2)
ELSE ROUND(payment.total, 2) END AS 'total',
答案 0 :(得分：2)
删除HAVING 1=1;优化程序可能不够智能而无法忽略它。请提供EXPLAIN SELECT（ not in html ）以查看优化工具的功能。
在这种情况下，复合PK似乎是错误的：PRIMARY KEY (id, payment_type_id)。请说明理由。
请解释status的含义或DOUBLE的需要：status DOUBLE
需要花一些精力来弄清楚查询为何如此缓慢。让我们首先抛出标准化部分，例如日期和事件名称和货币。这可以减少查询到足以找到所需的行，但不是每行的详细信息。如果它仍然很慢，让我们调试一下。如果它是“快速”，则逐个添加其他内容，以找出导致性能问题的原因。
每张桌子只有id PRIMARY KEY吗？或者是否有更多例外（例如payment）？
为question.var指定值似乎“错误”，但是后来使用LEFT暗示它是可选的。除非我在这个问题上有误，否则请将所有LEFT JOINs更改为INNER JOINs。
是否有任何表（可能是submission_entry和event_date_product）“多对多”映射表？如果是这样，请按照提示here获得一些性能提升。
当您回来时，请为每张桌子提供SHOW CREATE TABLE。
答案 1 :(得分：2)
遵循以下策略，
payment置于顶部 - 因为这似乎是最具决定性的我提供了您的查询的修订版本：
-- -----------------------------------------------------------------------------
-- Summarization of order_item
-- -----------------------------------------------------------------------------
drop temporary table if exists _ord_itm_sub_tot;
create temporary table _ord_itm_sub_tot(
primary key (payment_id)
)
SELECT
payment_id,
--
COUNT(
DISTINCT
CASE
WHEN(
`order_item`.status > 0 OR
(
`order_item`.status != -1 AND
`order_item`.status >= -2 AND
`payment`.payment_type_id != 8 AND
payment.make_order_free = 1
)
) THEN `order_item`.id
ELSE NULL
END
) AS qty,
--
SUM(order_item.sub_total) sub_total
FROM
order_item
inner join payment
on payment.id = order_item.payment_id
where order_item.status > 0.0 OR order_item.status = -2.0
group by payment_id;
-- -----------------------------------------------------------------------------
-- Summarization of payment_refunds
-- -----------------------------------------------------------------------------
drop temporary table if exists _pay_ref_tot;
create temporary table _pay_ref_tot(
primary key(payment_id)
)
SELECT
payment_refunds.payment_id AS `payment_id`,
sum(`payment_refund`.total) AS `refunds_total`
FROM
`payment_refunds`
INNER JOIN `payment` AS `payment_refund`
ON `payment_refund`.id = `payment_refunds`.payment_id_refund
GROUP BY `payment_refunds`.payment_id;
-- -----------------------------------------------------------------------------
-- Summarization of submission_entry
-- -----------------------------------------------------------------------------
drop temporary table if exists _sub_ent;
create temporary table _sub_ent(
primary key(form_submission_id)
)
select
submission_entry.form_submission_id,
GROUP_CONCAT(
DISTINCT (
CASE WHEN coalesce(submission_entry.text, '') THEN ' '
ELSE submission_entry.text
END
)
ORDER BY question.var
DESC SEPARATOR 0x1D
) AS buyer
from
submission_entry
LEFT JOIN question
ON(
question.id = submission_entry.question_id
AND question.var IN ('name', 'email')
)
group by submission_entry.form_submission_id;
-- -----------------------------------------------------------------------------
-- The result
-- -----------------------------------------------------------------------------
SELECT SQL_NO_CACHE
`payment`.`id` AS id,
`order_item`.`order_id` AS order_id,
--
_sub_ent.buyer,
--
event.name AS event,
--
_ord_itm_sub_tot.qty,
--
payment.currency AS `currency`,
--
_ord_itm_sub_tot.sub_total,
--
CASE
WHEN payment.make_order_free = 1 THEN ROUND(payment.total + COALESCE(refunds_total, 0), 2)
ELSE ROUND(payment.total, 2)
END AS 'total',
--
`payment_type`.`name` AS payment_type,
`payment_status`.`name` AS status,
`payment_status`.`id` AS status_id,
--
DATE_FORMAT(
CONVERT_TZ(order_item.`created`, '+0:00', '-8:00'),
'%Y-%m-%d %H:%i'
) AS 'created',
--
`user`.`name` AS 'agent',
event.id AS event_id,
payment.checked,
--
DATE_FORMAT(CONVERT_TZ(payment.checked_date, '+0:00', '-8:00'), '%Y-%m-%d %H:%i') AS checked_date,
DATE_FORMAT(CONVERT_TZ(payment.complete_date, '+0:00', '-8:00'), '%Y-%m-%d %H:%i') AS `complete date`,
--
`payment`.`delivery_status` AS `delivered`
FROM
`payment`
INNER JOIN(
`order_item`
INNER JOIN event
ON event.id = order_item.event_id
)
ON `order_item`.`payment_id` = payment.id
--
inner join _ord_itm_sub_tot
on _ord_itm_sub_tot.payment_id = payment.id
--
LEFT JOIN _pay_ref_tot
on _pay_ref_tot.payment_id = `payment`.id
--
INNER JOIN payment_status ON payment_status.id = payment.status
INNER JOIN payment_type ON payment_type.id = payment.payment_type_id
LEFT JOIN user ON user.id = payment.completed_by
--
LEFT JOIN _sub_ent
on _sub_ent.form_submission_id = `payment`.`form_submission_id`
WHERE
1 = 1
AND (payment.status > 0.0 OR payment.status = -3.0)
AND (order_item.status > 0.0 OR order_item.status = -2.0)
ORDER BY `order_item`.`order_id` DESC
LIMIT 10
来自你问题的查询显示了没有明确分组的聚合函数......这非常尴尬，在我的解决方案中，我尝试设计“有意义”的聚合。
请运行此版本并告诉我们您的发现。
请不要只关注正在运行的统计信息，还要注意汇总结果。
答案 2 :(得分：0)
（表格和查询对我来说太复杂了，无法为你做转换。但这里是步骤。）
refunds的情况下重新制定查询。也就是说，删除派生表并在复杂CASE中提及它。GROUP BY order_item ORDER BY order_item DESC LIMIT 10并执行已建议的任何其他优化。特别是，摆脱HAVING 1=1，因为它可能是一种可能的优化方式。类似的东西：
SELECT lots of stuff
FROM ( query from step 2 ) AS step2
LEFT JOIN ( ... ) AS refunds ON step2... = refunds...
ORDER BY step2.order_item DESC
重复ORDER BY，但GROUP BY和LIMIT都不需要重复。
为什么呢？这里的原则是......
目前，它会进入refunds相关子查询数千次，但只能丢掉10次。重新制定只能将其减少到10次。
（警告：我可能已经错过了一个微妙的事情，阻止了这个重新制定工作，因为我提出它。如果它不起作用，看看你是否可以让'原则'帮助你。）
答案 3 :(得分：0)
以下是每次看到包含大量连接和分页的查询时应该执行的最小值：您应该从第一个表（order_item）中选择那些以最小连接数分组的10个（LIMIT 10）ID可能然后将id连接回第一个表并进行所有其他连接。这样，您就不会在临时表中移动所有数千个不需要显示的列和行。
您可以查看内部联接和WHERE条件，GROUP BY和ORDER BY，以查看是否需要任何其他表来从第一个表中筛选出行，组或订单ID。在您的情况下，除了payment之外，您似乎不需要任何联接。
现在编写查询以选择这些ID：
SELECT o.order_id, o.payment_id
FROM order_item o
JOIN payment p
ON p.id = o.payment_id AND (p.status > 0.0 OR p.status = -3.0)
WHERE order_item.status > 0.0 OR order_item.status = -2.0
ORDER BY order_id DESC
LIMIT 10
如果单个订单可能有多笔付款，则应使用GROUP BY order_id DESC代替ORDER BY。为了使查询更快地运行，您需要在status列的order_item列上使用BTREE索引，甚至在(status, payment_id)上使用复合索引。
现在，当您确定ID是您期望的那些时，您可以进行所有其他连接：
SELECT order_item.order_id,
`payment`.`id`,
GROUP_CONCAT ... -- and so on from the original query
FROM (
SELECT o.order_id, o.payment_id
FROM order_item o
JOIN payment p
ON p.id = o.payment_id AND (p.status > 0.0 OR p.status = -3.0)
WHERE order_item.status > 0.0 OR order_item.status = -2.0
ORDER BY order_id DESC
LIMIT 10
) as ids
JOIN order_item ON ids.order_id = order_item.order_id
JOIN payment ON ids.payment_id = payment.id
LEFT JOIN ( ... -- and so on
这个想法是你大大降低了你需要处理的临时表。现在，连接选择的每一行都将用在结果集中。
UPD1：另一件事是你应该简化LEFT JOIN中的聚合：
SELECT
sum(payment.total) AS `refunds_total`,
refs.payment_id AS `payment_id`
FROM payment_refunds refs
JOIN payment ON payment.id = refs.payment_id_refund
GROUP BY refs.payment_id
甚至用相关的子查询替换LEFT JOIN，因为只对那10行执行相关（确保，你使用这三个列的整个查询作为子查询，否则，将为每个子列计算相关性在GROUP BY之前的结果连接中的行：
SELECT
ids.order_id,
ids.payment_id,
(SELECT SUM(p.total)
FROM payment_refunds refs
JOIN payment p
ON refs.payment_id_refund = p.id
WHERE refs.payment_id = ids.payment_id
) as refunds_total
FROM (
SELECT o.order_id, o.payment_id
FROM order_item o
JOIN payment p
ON p.id = o.payment_id AND (p.status > 0.0 OR p.status = -3.0)
WHERE order_item.status > 0.0 OR order_item.status = -2.0
ORDER BY order_id DESC
LIMIT 10
) as ids
您还需要(payment_id, payment_id_refund)上的索引payment_refunds，您甚至可以在付款时尝试覆盖索引(payment_id, total)。