我有一个查询基本上结合了动作表并按时间顺序从中选择,同时保留了分页。
有更高效/更好的方法吗?查询需要3秒钟。并不可怕..但我认为还有改进的余地,我会用它很多..
谢谢!
SELECT
`newsletters_subscribers`.`email`,
`newsletters_subscribers`.`first_name`,
`newsletters_subscribers`.`last_name`,
`newsletters_subscribers`.`id` AS subscriber_id,
COUNT(DISTINCT newsletters_opens.id) AS opens,
COUNT(DISTINCT newsletters_clicks.id) AS clicks,
COUNT(DISTINCT newsletters_forwards.id) AS forwards
FROM `thebookrackqccom_newsletters_subscribers` newsletters_subscribers
LEFT JOIN
`thebookrackqccom_newsletters_opens` newsletters_opens
ON `newsletters_opens`.`subscriber_id` = `newsletters_subscribers`.`id`
AND newsletters_opens.newsletter_id = 1
LEFT JOIN
`thebookrackqccom_newsletters_clicks` newsletters_clicks
ON `newsletters_clicks`.`subscriber_id` = `newsletters_subscribers`.`id`
AND newsletters_clicks.newsletter_id = 1
LEFT JOIN
`thebookrackqccom_newsletters_forwards` newsletters_forwards
ON `newsletters_forwards`.`subscriber_id` = `newsletters_subscribers`.`id`
AND newsletters_forwards.newsletter_id = 1
WHERE
( newsletters_opens.id IS NOT NULL
OR newsletters_clicks.id IS NOT NULL
OR newsletters_forwards.id IS NOT NULL )
GROUP BY
`newsletters_subscribers`.`id`
ORDER BY
`newsletters_subscribers`.`email` ASC
LIMIT 25
答案 0 :(得分:1)
您需要的是查询可以使用的索引。在这三个表中的每一个表上(newsletter_id, subscribe_id)
上的复合索引都会有所帮助。
您也可以像这样重写查询:
SELECT
s.email,
s.first_name,
s.last_name,
s.id AS subscriber_id,
COALESCE(o.opens, 0) AS opens,
COALESCE(c.clicks, 0) AS clicks,
COALESCE(f.forwards, 0) AS forwards
FROM thebookrackqccom_newsletters_subscribers AS s
LEFT JOIN
( SELECT subscriber_id,
COUNT(*) AS opens
FROM thebookrackqccom_newsletters_opens
WHERE newsletters_opens.newsletter_id = 1
) AS o ON o.subscriber_id = s.id
LEFT JOIN
( SELECT subscriber_id,
COUNT(*) AS clicks
FROM thebookrackqccom_newsletters_clicks
WHERE newsletter_id = 1
) AS c ON c.subscriber_id = s.id
LEFT JOIN
( SELECT subscriber_id,
COUNT(*) AS forwards
FROM thebookrackqccom_newsletters_forwards
WHERE newsletter_id = 1
) AS f ON f.subscriber_id = s.id
WHERE ( o.subscriber_id IS NOT NULL
OR c.subscriber_id IS NOT NULL
OR f.subscriber_id IS NOT NULL )
ORDER BY
s.email ASC
LIMIT 25
答案 1 :(得分:0)
尝试此查询我希望您获得更好的执行时间
QUERY
SELECT
`newsletters_subscribers`.`email`,
`newsletters_subscribers`.`first_name`,
`newsletters_subscribers`.`last_name`,
`newsletters_subscribers`.`id` AS subscriber_id,
@nopen := coalesce( N_OPEN.NOPENIDCOUNT, 000000 ) as opens,
@nclick := coalesce( N_CLICK.NCLICKIDCOUNT, 000000 ) as clicks,
@nfwd := coalesce( N_FWD.NFWDIDCOUNT, 000000 ) as forwards
FROM
(select @nopen := 0,@nclick := 0,@nfwd :=0) sqlvars,
`thebookrackqccom_newsletters_subscribers` AS newsletters_subscribers
LEFT JOIN (SELECT `newsletters_opens`.`subscriber_id`,
COUNT(newsletters_opens.id) AS NOPENIDCOUNT
FROM `thebookrackqccom_newsletters_opens` AS newsletters_opens
WHERE newsletters_opens.newsletter_id = 1) AS N_OPEN
ON N_OPEN.subscriber_id = `newsletters_subscribers`.`id`
LEFT JOIN (SELECT `newsletters_clicks`.`subscriber_id`,
COUNT(newsletters_clicks.id) AS NCLICKIDCOUNT
FROM `thebookrackqccom_newsletters_clicks` AS newsletters_clicks
WHERE newsletters_clicks.newsletter_id = 1) AS N_CLICK
ON N_CLICK.subscriber_id = `newsletters_subscribers`.`id`
LEFT JOIN (SELECT `newsletters_forwards`.`subscriber_id`,
COUNT(newsletters_forwards.id) AS NFWDIDCOUNT
FROM `thebookrackqccom_newsletters_forwards` AS newsletters_forwards
WHERE newsletters_forwards.newsletter_id = 1) AS N_FWD
ON N_FWD.subscriber_id = `newsletters_subscribers`.`id`
GROUP BY `newsletters_subscribers`.`id`
ORDER BY `newsletters_subscribers`.`email` ASC
LIMIT 25