Question

我的数据库中有三个表 - APPLICATION，APPLICANT和ADDRESS

APPLICATION中有一行。

APPLICANT可以通过APPLICATION将1或2行链接回APPLICATION_ID。

ADDRESS可以通过APPLICANT将1行，2行或3行链接回APPLICANT_ID。

APPLICATION - ＆gt; （APPLICATION_ID上的1对多） - ＆gt; APPLICANT - ＆gt; （APPLICANT_ID上的1对多） - ＆gt; ADDRESS

我需要编写一个查询，从每个表中提取特定字段（从＆＃39;所有信息＆＃39;中更改）到1个结果集。结果需要包含一个结果行中每个应用程序的所有可能信息。有人可以请我指出最佳解决方案吗？

我希望问题很清楚。我已经搜索了SO，但只能找到一些特定于案例的答案，并且没有关于1对多连接的一般说法。

好吧，我认为我应该做一点帮助，以帮助那些真正花时间思考这个问题的人。这是来自所有三个表的一些示例虚拟数据。

APPLICATION
-----------
APPLICATION_ID|APP1|APP2|OTHER_STUFF
1             |1   |1   |x

APPLICANT
---------
APPLICANT_ID|APPLICATION_ID|FORENAME|OTHER_STUFF
1           |1             |Homer   |x
2           |1             |Marge   |x

ADDRESS
-------
ADDRESS_ID|APPLICANT_ID|STREET           |OTHER_STUFF
1         |1           |Sesame Street    |x
2         |1           |Evergreen Terrace|x
3         |2           |Evergreen Terrace|x

SQL查询的结果看起来像这样（希望如此）;

APPLICATION_ID|APPLICANT_ID1|FORENAME1|ADDRESS_ID1|STREET1      |ADDRESS_ID2|STREET2          |APPLICANT_ID2|FORENAME_2|ADDRESS_ID3|STREET3
1             |1            |Homer    |1          |Sesame Street|2          |Evergreen Terrace|2            |Marge     |3          |Evergreen Terrace

由于

Answer 1

您一定会使用以下查询。希望这会对你有所帮助。

SELECT * FROM APPLICATION as App

INNER JOIN APPLICANT as A1 on A1.APPLICATION_ID = App.APPLICATION_ID

INNER JOIN ADDRESS as A2 on A2.APPLICANT_ID = A1.APPLICANT_ID

Answer 2

; WITH applicants AS (
  SELECT applicant_id
       , application_id
       , forename
       , other_stuff
       , Row_Number() OVER (PARTITION BY application_id ORDER BY applicant_id) As sequence
  FROM   applicant
)
, addresses AS (
  SELECT address_id
       , applicant_id
       , street
       , other_stuff
       , Row_Number() OVER (PARTITION BY applicant_id ORDER BY address_id) As sequence
  FROM   address
)
SELECT application.application_id
     , first_applicants.applicant_id As applicant_id1
     , first_applicants.forename As forename1
     , first_applicants_first_addresses.address_id As address_id1
     , first_applicants_first_addresses.street As street1
     , first_applicants_second_addresses.address_id As address_id2
     , first_applicants_second_addresses.street As street2
     , second_applicants.applicant_id As applicant_id2
     , second_applicants.forename As forename2
     , second_applicants_first_addresses.address_id As address_id3
     , second_applicants_first_addresses.street As street3
     , second_applicants_second_addresses.address_id As address_id4
     , second_applicants_second_addresses.street As street4     
FROM   application
 LEFT
  JOIN applicants As first_applicants
    ON first_applicants.application_id = application.application_id
   AND first_applicants.sequence = 1
 LEFT
  JOIN addresses As first_applicants_first_addresses
    ON first_applicants_first_addresses.applicant_id = first_applicants.applicant_id
   AND first_applicants_first_addresses.sequence = 1
 LEFT
  JOIN addresses As first_applicants_second_addresses
    ON first_applicants_second_addresses.applicant_id = first_applicants.applicant_id
   AND first_applicants_second_addresses.sequence = 2
 LEFT
  JOIN applicants As second_applicants
    ON second_applicants.application_id = application.application_id
   AND second_applicants.sequence = 2
 LEFT
  JOIN addresses As second_applicants_first_addresses
    ON second_applicants_first_addresses.applicant_id = second_applicants.applicant_id
   AND second_applicants_first_addresses.sequence = 1
 LEFT
  JOIN addresses As second_applicants_second_addresses
    ON second_applicants_second_addresses.applicant_id = second_applicants.applicant_id
   AND second_applicants_second_addresses.sequence = 2
WHERE  application.application_id = 1
;

Answer 3

select APPLICATION.*, APPLICANT.*, ADDRESS.* 
  from APPLICATION
  join APPLICANT 
    on APPLICATION.APPLICATION_ID = APPLICANT.APPLICATION_ID
  join ADDRESS
    on APPLICANT.APPLICATION_ID = APPLICANT.APPLICATION_ID

一对多加入

3 个答案: