将 2 个内部连接结果合并为一个结果
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【中文标题】将 2 个内部连接结果合并为一个结果【英文标题】：Join 2 inner join result into one result 【发布时间】：2017-12-22 00:05:54 【问题描述】：大家我有问题。我有两个单独运行良好的查询。但我想提出一个我尝试过的查询，但这些都行不通。我需要加入两个输出。
第一次查询：
SELECT q_internal_table.q_part_id,
q_external_table.q_external_id,
q_external_table.q_external_approve,
q_external_table.q_order_id,
q_internal_table.q_internal_id,
q_internal_table.q_internal_approve
FROM (SELECT parts.id AS q_part_id,
parts.order_id AS q_order_id,
external_reports.id AS q_external_id,
external_reports.approved AS q_external_approve
FROM parts
INNER JOIN external_reports
ON( parts.id = external_reports.part_id ))
q_external_table
INNER JOIN (SELECT parts.id AS q_part_id,
internal_reports.id AS q_internal_id,
internal_reports.approved AS q_internal_approve
FROM parts
INNER JOIN internal_reports
ON( parts.id = internal_reports.part_id ))
q_internal_table
ON( q_external_table.q_part_id = q_internal_table.q_part_id )
WHERE ( q_external_table.q_external_approve = 'Y'
OR q_internal_table.q_internal_approve = 'Y' )
第二次查询：
SELECT q_five_internal_table.q_five_part_id,
q_five_internal_table.q_five_internal_id,
q_five_internal_table.q_five_internal_approve,
q_five_external_table.q_five_external_id,
q_five_external_table.q_five_external_approve,
q_five_external_table.q_five_order_id
FROM (SELECT parts.id AS q_five_part_id,
parts.order_id AS q_five_order_id,
five_way_external_reports.id AS q_five_external_id,
five_way_external_reports.approved AS q_five_external_approve
FROM parts
INNER JOIN five_way_external_reports
ON( parts.id = five_way_external_reports.part_id ))
q_five_external_table
INNER JOIN (SELECT parts.id AS q_five_part_id,
five_way_internal_reports.id AS q_five_internal_id,
five_way_internal_reports.approved AS q_five_internal_approve
FROM parts
INNER JOIN five_way_internal_reports
ON( parts.id = five_way_internal_reports.part_id ))
q_five_internal_table
ON( q_five_external_table.q_five_part_id = q_five_internal_table.q_five_part_id )
WHERE ( q_five_external_table.q_five_external_approve = 'Y'
OR q_five_internal_table.q_five_internal_approve = 'Y' )
第一次查询结果：
第二次查询结果：
我试过这个查询：-
SELECT q_internal_external_table.*,
q_five_internal_external_table.*
FROM (SELECT q_internal_table.q_part_id,
q_external_table.q_external_id,
q_external_table.q_external_approve,
q_external_table.q_order_id,
q_internal_table.q_internal_id,
q_internal_table.q_internal_approve
FROM (SELECT parts.id AS q_part_id,
parts.order_id AS q_order_id,
external_reports.id AS q_external_id,
external_reports.approved AS q_external_approve
FROM parts
INNER JOIN external_reports
ON( parts.id = external_reports.part_id ))
q_external_table
INNER JOIN (SELECT parts.id AS q_part_id,
internal_reports.id AS q_internal_id,
internal_reports.approved AS q_internal_approve
FROM parts
INNER JOIN internal_reports
ON( parts.id = internal_reports.part_id ))
q_internal_table
ON( q_external_table.q_part_id = q_internal_table.q_part_id )
WHERE ( q_external_table.q_external_approve = 'Y'
OR q_internal_table.q_internal_approve = 'Y' ))
q_internal_external_table
INNER JOIN (SELECT q_five_internal_table.q_five_part_id,
q_five_internal_table.q_five_internal_id,
q_five_internal_table.q_five_internal_approve,
q_five_external_table.q_five_order_id,
q_five_external_table.q_five_external_id,
q_five_external_table.q_five_external_approve
FROM (SELECT parts.id AS q_five_part_id,
parts.order_id AS
q_five_order_id
,
five_way_external_reports.id AS
q_five_external_id,
five_way_external_reports.approved AS
q_five_external_approve
FROM parts
INNER JOIN five_way_external_reports
ON( parts.id = five_way_external_reports.part_id ))
q_five_external_table
INNER JOIN (SELECT parts.id AS q_five_part_id,
five_way_internal_reports.id AS
q_five_internal_id,
five_way_internal_reports.approved AS
q_five_internal_approve
FROM parts
INNER JOIN five_way_internal_reports
ON( parts.id = five_way_internal_reports.part_id ))
q_five_internal_table
ON ( q_five_external_table.q_five_part_id =
q_five_internal_table.q_five_part_id )
WHERE ( q_five_external_table.q_five_external_approve = 'Y'
OR q_five_internal_table.q_five_internal_approve = 'Y' ))
q_five_internal_external_table
ON ( q_internal_external_table.q_part_id =
q_five_internal_external_table.q_five_part_id )
查询结果：-
在这个查询中有三个结果。但我希望 7 结果与第一个查询结果一样。三个答案已经在第一个查询结果中。我也需要休息4。我认为问题出在ON q_internal_external_table.q_part_id = q_five_internal_external_table.q_five_part_id。但是不知道怎么解决。
请帮助我。提前致谢。
【问题讨论】：
1) 预期的结果是什么？ 2) 你尝试过什么，这些尝试出了什么问题？ 你的意思是 UNION - 谷歌 @Shadow：我更新了问题。我想加入两个答案。我还添加了我尝试过的 SQL。 类似select * from (1st select) as t1 join (2nd select) as t2 on t1.q_part_id = t2.q_five_part_id.
加入意味着什么？请演示使用问题中已有的 7 + 3 记录。
【参考方案1】：
特别感谢@dnoeth。好人。
修改后的SQL是，
select q_internal_external_table.*,q_five_internal_external_table.* from
(select q_internal_table.q_part_id,
q_external_table.q_external_id,q_external_table.q_external_approve,
q_external_table.q_order_id,q_internal_table.q_internal_id,
q_internal_table.q_internal_approve from(SELECT parts.id as
q_part_id,parts.order_id as q_order_id,external_reports.id as
q_external_id,external_reports.approved as q_external_approve from parts
LEFT join external_reports on(parts.id=external_reports.part_id))
q_external_table LEFT join (SELECT parts.id as q_part_id,internal_reports.id
as q_internal_id,internal_reports.approved as q_internal_approve from parts
LEFT join internal_reports on(parts.id=internal_reports.part_id))
q_internal_table on(q_external_table.q_part_id=q_internal_table.q_part_id)
where (q_external_table.q_external_approve='Y' or
q_internal_table.q_internal_approve='Y')) q_internal_external_table LEFT
join (select q_five_internal_table.q_five_part_id,
q_five_internal_table.q_five_internal_id,
q_five_internal_table.q_five_internal_approve,
q_five_external_table.q_five_order_id,
q_five_external_table.q_five_external_id,
q_five_external_table.q_five_external_approve from (SELECT parts.id as
q_five_part_id,parts.order_id as q_five_order_id,
five_way_external_reports.id as q_five_external_id,
five_way_external_reports.approved as q_five_external_approve from parts
LEFT join five_way_external_reports on(parts.id=
five_way_external_reports.part_id)) q_five_external_table LEFT join (SELECT
parts.id as q_five_part_id,five_way_internal_reports.id as
q_five_internal_id,five_way_internal_reports.approved as
q_five_internal_approve from parts LEFT join five_way_internal_reports
on(parts.id=five_way_internal_reports.part_id)) q_five_internal_table on
(q_five_external_table.q_five_part_id=q_five_internal_table.q_five_part_id)
where (q_five_external_table.q_five_external_approve='Y' or
q_five_internal_table.q_five_internal_approve='Y'))
q_five_internal_external_table on (q_internal_external_table.q_part_id
=q_five_internal_external_table.q_five_part_id)
答案：（我想要的）
【讨论】：
【参考方案2】：如果我们假设您查询的最终目标是在下表中找到一个parts.id，那么您更新后的查询似乎是正确的（语义上）： external_reports，internal_reports，five_way_external_reports，five_way_internal_reports
在合并查询1和查询2的行时使用联合。
但是，如果您尝试查找匹配的列（连接子句），则应使用 JOIN 而不是 UNION。
另外，请尝试在您的查询中添加 EXPLAIN 和 EXPLAIN EXTENDED 前缀以了解查询计划。
【讨论】：
【参考方案3】：你需要的是UNION。
两个查询需要具有相同数量的列和类型。在您的情况下，您需要更改所选列的顺序以使UNION。
应该是这样的：
SELECT q_internal_table.q_part_id,
q_external_table.q_external_id,
q_external_table.q_external_approve,
q_external_table.q_order_id,
q_internal_table.q_internal_id,
q_internal_table.q_internal_approve
FROM (SELECT parts.id AS q_part_id,
parts.order_id AS q_order_id,
external_reports.id AS q_external_id,
external_reports.approved AS q_external_approve
FROM parts
INNER JOIN external_reports
ON( parts.id = external_reports.part_id ))
q_external_table
INNER JOIN (SELECT parts.id AS q_part_id,
internal_reports.id AS q_internal_id,
internal_reports.approved AS q_internal_approve
FROM parts
INNER JOIN internal_reports
ON( parts.id = internal_reports.part_id ))
q_internal_table
ON( q_external_table.q_part_id = q_internal_table.q_part_id )
WHERE ( q_external_table.q_external_approve = 'Y'
OR q_internal_table.q_internal_approve = 'Y' )
UNION ALL
SELECT q_five_internal_table.q_five_part_id,
q_five_external_table.q_five_external_id,
q_five_external_table.q_five_external_approve,
q_five_external_table.q_five_order_id,
q_five_internal_table.q_five_internal_id,
q_five_internal_table.q_five_internal_approve
FROM (SELECT parts.id AS q_five_part_id,
parts.order_id AS q_five_order_id,
five_way_external_reports.id AS q_five_external_id,
five_way_external_reports.approved AS q_five_external_approve
FROM parts
INNER JOIN five_way_external_reports
ON( parts.id = five_way_external_reports.part_id ))
q_five_external_table
INNER JOIN (SELECT parts.id AS q_five_part_id,
five_way_internal_reports.id AS q_five_internal_id,
five_way_internal_reports.approved AS q_five_internal_approve
FROM parts
INNER JOIN five_way_internal_reports
ON( parts.id = five_way_internal_reports.part_id ))
q_five_internal_table
ON( q_five_external_table.q_five_part_id = q_five_internal_table.q_five_part_id )
WHERE ( q_five_external_table.q_five_external_approve = 'Y'
OR q_five_internal_table.q_five_internal_approve = 'Y' )
请注意，我已更改第二个查询中列的顺序以匹配第一个查询的类型。
【讨论】：
即使知道预期的输出也可能是UNION 或 EXCEPT 或 INTERSECT :-)
感谢您的帮助。但是在答案中没有显示第二个查询结果。以上是关于将 2 个内部连接结果合并为一个结果的主要内容，如果未能解决你的问题，请参考以下文章
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