我有一个mysql表，我试图运行单个/ 嵌套查询。
SELECT
`subscriber_id`,
(SELECT...?) AS total_campaigns,
(SELECT...?) AS total_opens,
(total_opens / total_campaigns) as percentage
FROM
`campaigns_table`
值为2的type列显示广告系列已打开。
我希望返回包含所有subscriber_id的结果集，其中包含所有广告系列和总计广告
+---------------+-----------------+-------------+------------+
| subscriber_id | total_campaigns | total_opens | percentage |
+---------------+-----------------+-------------+------------+
| 1 | 2 | 1 | 0.5 |
| 2 | 2 | 2 | 1.0 |
| 3 | 2 | 0 | 0.0 |
| 4 | 1 | 0 | 0.0 |
| 5 | 1 | 1 | 1.0 |
+---------------+-----------------+-------------+------------+
subscriber_id 1将是total_campaigns = 2（campaign_id＆＃39; 37428,37239）和total_opens = 1（仅campaign_id 27428具有类型2记录）
+---------------+-------------+------------+-------+------+---------+
| subscriber_id | campaign_id | timestamp | count | type | link_id |
+---------------+-------------+------------+-------+------+---------+
| 1 | 37428 | 1434513738 | 1 | 1 | 0 |
| 1 | 37428 | 1434513758 | 1 | 1 | 0 |
| 1 | 37428 | 1434513338 | 2 | 2 | 0 |
| 1 | 37429 | 1434511738 | 1 | 1 | 0 |
| 1 | 37429 | 1434311738 | 1 | 1 | 1 |
| 2 | 37428 | 1534513738 | 1 | 1 | 0 |
| 2 | 37428 | 1534513758 | 1 | 1 | 0 |
| 2 | 37428 | 1534513338 | 2 | 2 | 0 |
| 2 | 37429 | 1534511738 | 1 | 1 | 1 |
| 2 | 37429 | 1534311738 | 1 | 2 | 0 |
| 3 | 37428 | 1534513738 | 1 | 1 | 0 |
| 3 | 37429 | 1534511738 | 1 | 1 | 1 |
| 4 | 57428 | 1534513738 | 1 | 1 | 0 |
| 4 | 57428 | 1534513758 | 1 | 1 | 0 |
| 5 | 57428 | 1534513338 | 3 | 2 | 0 |
+---------------+-------------+------------+-------+------+---------+
使用下面@ spencer7593的答案。然后我如何使用结果更新另一个表？
尝试做这样的事情（我的工作方式并不好）
UPDATE `subscribers` a
LEFT JOIN `campaigns_table` b ON a.`ID` = b.`subscriber_id`
SET a.`STATUS` = 2
FROM(
SELECT t.subscriber_id
, COUNT(DISTINCT t.campaign_id) AS total_campaigns
, COUNT(DISTINCT IF(t.type=2,t.campaign_id,NULL)) AS open_campaigns
, COUNT(DISTINCT IF(t.type=2,t.campaign_id,NULL))
/ COUNT(DISTINCT t.campaign_id) AS percentage
FROM `campaigns_table` t
GROUP BY t.subscriber_id
HAVING COUNT(DISTINCT t.campaign_id) > 5 AND COUNT(DISTINCT IF(t.type=2,t.campaign_id,NULL)) = 0
ORDER BY t.subscriber_id
) i
答案 0 :(得分：0)
在我看来，这是一个可以通过条件聚合解决的问题，只需一次通过表，不需要任何嵌套查询。
SELECT t.subscriber_id
, COUNT(DISTINCT t.campaign_id) AS total_campaigns
, COUNT(DISTINCT IF(t.type=2,t.campaign_id,NULL)) AS open_campaigns
, COUNT(DISTINCT IF(t.type=2,t.campaign_id,NULL))
/ COUNT(DISTINCT t.campaign_id) AS percentage
FROM `campaigns_table` t
GROUP BY t.subscriber_id
ORDER BY t.subscriber_id
可以使用嵌套查询生成等效结果，但一般来说，单次通过表（使用单个SELECT）将具有更高效的访问计划。
如果需要使用嵌套查询（我不明白为什么会这样），一般来说，我更喜欢在SELECT列表中使用相关查询的内联视图。
SELECT q.subscriber_id
, SUM(1) AS total_campaigns
, SUM(q.open_campaign) AS open_campaigns
, SUM(q.open_campaign)/SUM(1) AS percentage
FROM ( SELECT t.subscriber_id
, t.campaign
, MAX(t.type=2) AS `open_campaign`
FROM `campaigns_table` t
WHERE t.campaign IS NOT NULL
GROUP
BY t.subscriber_id
, t.campaign
) q
ORDER BY q.subscriber
如果我们想在SELECT列表中使用嵌套查询，那么它们就是相关的子查询......
SELECT s.subscriber
, ( SELECT COUNT(DISTINCT t.campaign)
FROM `campaigns_table` t
WHERE t.subscriber = s.subscriber
) AS total_campaigns
, ( SELECT COUNT(DISTINCT o.campaign)
FROM `campaigns_table` o
WHERE o.subscriber = s.subscriber
AND o.type=2
) AS open_campaigns
, ( SELECT COUNT(DISTINCT po.campaign)
FROM `campaigns_table` po
WHERE po.subscriber = s.subscriber
AND po.type=2
)
/ ( SELECT COUNT(DISTINCT pt.campaign)
FROM `campaigns_table` pt
WHERE pt.subscriber = s.subscriber
) AS percentage
FROM ( SELECT r.subscriber
FROM `campaigns_table` r
GROUP BY r.subscriber
) s
ORDER BY s.subscriber