[英]SQL Left Join Multiple Tables and count values conditionally in each table
我在将SQL语句正确组合时遇到了一些麻烦,因为我没有太多经验SQL,尤其是聚合函数。 安全地说,我真的不知道我在做什么基本的SQL结构之外。 我可以做常规连接,但不能复杂。
我有一些表: '调查' , '问题' , '会话' , 'ParentSurvey'和'ParentSurveyQuestion' 。 在结构上,调查可以有问题,可以让用户开始调查(会话),并且可以进行父调查,其问题可以导入到当前调查中。
我想要做的是在调查表中获取每个调查的信息; 它有多少问题,已开始的会议次数(有条件地,尚未完成的会议),以及父母调查中的问题数量。 三个连接的表可以但不必包含任何值,如果它们不包含,那么应该通过COUNT返回0。 三个表中的公共字段是'survey_id'的变体
到目前为止,这是我的SQL,我将表结构放在它下面。
SELECT
`kp_survey_id`,
COALESCE( q.cnt, 0 ) AS questionsAmount,
COALESCE( s.cnt, 0 ) AS sessionsAmount
COALESCE( p.cnt, 0 ) AS parentQAmount,
FROM `Survey`
LEFT JOIN <-- I'd like the count of questions for this survey
( SELECT COUNT(*) AS cnt
FROM Questions
GROUP BY kf_survey_id ) q
ON Survey.kp_survey_id = Questions.kf_survey_id
LEFT JOIN
( SELECT COUNT(*) AS cnt <-- I'd like the count of started sessions for this survey
FROM Session
WHERE session_status = 'started' <-- should this be Session.session_status?
GROUP BY kf_survey_id ) s
ON Survey.kp_survey_id = Session.kf_survey_id
LEFT JOIN
( SELECT COUNT(*) AS cnt <-- I'd like the count of questions in the parent survey with this survey id
FROM ParentSurvey
GROUP BY kp_parent_survey_id ) p
ON Survey.kf_parent_survey_id = ParentSurveyQuestion.kf_parent_survey_id
'kp'前缀表示主键,而'kf'前缀表示外键
结构体:
调查: 'kp_survey_id'| 'kf_parent_survey_id'
问题: 'kp_question_id'| 'kf_survey_id'
会话: 'kp_session_id'| 'kf_survey_id'| 'session_status'
ParentSurvey: 'kp_parent_survey_id'| 'survey_name'
ParentSurveyQuestion: 'kp_parent_question_id'| 'kf_parent_survey_id'
每个表中还有其他列,如'name'或'account_id',但我不认为它们在这种情况下很重要
我想知道我是否正确地做了这件事,或者我是否遗漏了什么。 我正在重新调整我在stackoverflow上找到的一些代码并修改它以满足我的需求,因为我还没有在这个网站上看到超过三个表的条件聚合。
我的预期输出是这样的:
kp_survey_id | questionsAmount | sessionsAmount | parentQAmount
1 | 3 | 0 | 3
2 | 0 | 5 | 3
你需要做的一件事是纠正你的联接。 当您加入子查询时,您需要使用子查询的别名。 在您的情况下,您正在使用子查询中使用的表的别名。
您需要更改的另一件事是在子查询中包含您希望在JOIN中使用的字段。
进行这些更改并尝试运行。 你得到错误或想要的结果吗?
SELECT
`kp_survey_id`,
COALESCE( q.cnt, 0 ) AS questionsAmount,
COALESCE( s.cnt, 0 ) AS sessionsAmount
COALESCE( p.cnt, 0 ) AS parentQAmount,
FROM `Survey`
LEFT JOIN <-- I'd like the count of questions for this survey
( SELECT kf_survey_id, COUNT(*) AS cnt
FROM Questions
GROUP BY kf_survey_id ) q
ON Survey.kp_survey_id = q.kf_survey_id
LEFT JOIN
( SELECT kf_survey_id, COUNT(*) AS cnt <-- I'd like the count of started sessions for this survey
FROM Session
WHERE session_status = 'started' <-- should this be Session.session_status?
GROUP BY kf_survey_id ) s
ON Survey.kp_survey_id = s.kf_survey_id
LEFT JOIN
( SELECT kp_parent_survey_id, COUNT(*) AS cnt <-- I'd like the count of questions in the parent survey with this survey id
FROM ParentSurvey
GROUP BY kp_parent_survey_id ) p
ON Survey.kf_parent_survey_id = p.kf_parent_survey_id
我认为你非常接近 - 只需要修复你的连接并在子查询中包含调查ID以用于这些连接:
SELECT
`kp_survey_id`,
COALESCE( q.cnt, 0 ) AS questionsAmount,
COALESCE( s.cnt, 0 ) AS sessionsAmount
COALESCE( p.cnt, 0 ) AS parentQAmount,
FROM `Survey`
LEFT JOIN
( SELECT COUNT(*) cnt, kf_survey_id AS cnt
FROM Questions
GROUP BY kf_survey_id ) q
ON Survey.kp_survey_id = q.kf_survey_id
LEFT JOIN
( SELECT COUNT(*) cnt, kf_survey_id
FROM Session
WHERE session_status = 'started'
GROUP BY kf_survey_id ) s
ON Survey.kp_survey_id = s.kf_survey_id
LEFT JOIN
( SELECT COUNT(*) cnt, kp_parent_survey_id
FROM ParentSurvey
GROUP BY kp_parent_survey_id ) p
ON Survey.kf_parent_survey_id = p.kp_parent_survey_id
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