我知道一个类似于此的问题已经发布但我无法通过尝试在谷歌上找到的东西找到我的SQL查询的问题
我得到的错误是“你的SQL语法错误”
这是我的代码
select
campaigns.campaign_name,
calls.count,
first.count as first,
second.count as second,
third.count as third,
fourth.count as fourth,
fifth.count as fifth,
sixth.count as sixth
from campaigns
LEFT JOIN
(
select count(*) as count, campaign_id FROM calls where direction = 'Incoming' group by campaign_id
) as calls
ON campaigns.id = calls.campaign_id
LEFT JOIN
(
select count(*) as count,
campaign_id FROM calls
where direction = 'Incoming'
AND DATE(created_at) BETWEEN '2014-01-01' AND '2016-01-01'
group by campaign_id
) as first
LEFT JOIN
(
select count(*) as count,
campaign_id FROM calls
where direction = 'Incoming'
AND DATE(created_at) BETWEEN '2014-01-01' AND '2016-01-01'
group by campaign_id
) as second
ON campaigns.id = second.campaign_id
LEFT JOIN
(
select count(*) as count,
campaign_id FROM calls
where direction = 'Incoming'
AND DATE(created_at) BETWEEN '2014-01-01' AND '2016-01-01'
group by campaign_id
) as third
ON campaigns.id = third.campaign_id
LEFT JOIN
(
select count(*) as count,
campaign_id FROM calls
where direction = 'Incoming'
AND DATE(created_at) BETWEEN '2014-01-01' AND '2016-01-01'
group by campaign_id
) as fourth
ON campaigns.id = fourth.campaign_id
LEFT JOIN
(
select count(*) as count,
campaign_id FROM calls
where direction = 'Incoming'
AND DATE(created_at) BETWEEN '2014-01-01' AND '2016-01-01'
group by campaign_id
) as fifth
ON campaigns.id = fifth.campaign_id
LEFT JOIN
(
select count(*) as count,
campaign_id FROM calls
where direction = 'Incoming'
AND DATE(created_at) BETWEEN '2014-01-01' AND '2016-01-01'
group by campaign_id
) as sixth
ON sixth.campaign_id = campaigns.id
group by campaigns.id
错误似乎在第69行
sql查询的最后一行ON sixth.campaign_id = campaigns.id
答案 0 :(得分:1)
没有理由有这么复杂的查询。只需使用条件聚合:
select c.campaign_name,
sum(direction = 'Incoming') as cnt,
sum(direction = 'Incoming' and DATE(created_at) BETWEEN '2014-01-01' AND '2016-01-01') as first,
. . .
from campaigns c left join
calls
on c.campaign_id = calls.campaign_id
group by c.campaign_name;
您的所有条件似乎都相同,但您可以为其他条件添加其他sum()
。