FULL OUTER JOIN 不返回所有表的内容

如何解决FULL OUTER JOIN 不返回所有表的内容

我有下面提到的 3 个表。

表格

BUG

Id | Name | Other information ...
1  | Bug1 | ...
2  | Bug2 | ...
3  | Bug3 | ...
4  | Bug4 | ...
5  | Bug5 | ...

链接

Id  | Test_id | Bug_id
100 | 1000    | 4
101 | 1100    | 2
102 | 1200    | 2
103 | 1200    | 5

测试

Id   | Name | Other information ...
1000 | TC1  | ...
1100 | TC2  | ...
1200 | TC3  | ...
1300 | TC4  | ...
1400 | TC5  | ...
1500 | TC6  | ...

请求

SELECT Bug.Id,Bug.Name,Test.Id,Test.Name
FROM Test FULL OUTER JOIN Link ON Link.Test_id = Test.Id
FULL OUTER JOIN Bug ON Link.Bug_id = Bug.Id
INNER JOIN Test_Detail ON Test_Detail.Test_id = Test.Id
INNER JOIN Release ON Release.Id = Test.Release_id
INNER JOIN Cycle ON Cycle.Release_id = Release.Id

然而，这并没有提供预期的结果......

预期结果

如果有链接，我想获取有关该错误的信息和测试
如果有没有链接的错误，我想得到错误信息
如果有没有链接的测试，我想获取有关测试的信息

结果

BugId | BugName | TestId | TestName
1     | Bug1    |
2     | Bug2    | 1100   | TC2
2     | Bug2    | 1200   | TC3
3     | Bug3    |
4     | Bug4    | 1000   | TC1
5     | Bug5    | 1200   | TC2
      |         | 1300   | TC4
      |         | 1400   | TC5
      |         | 1500   | TC6

结论

FULL OUTER JOIN 不是要使用的正确连接？你能帮助理解为什么它没有按预期工作吗？

谢谢

编辑 1

结果基于 Charlieface 的回答。

我收到了错误、测试和链接。但我只需要链接（如果存在）。如果没有链接，我需要测试和错误详细信息。如上文预期结果部分所述。

BugId | BugName | TestId | TestName
      |         | 1100   | TC2
      |         | 1200   | TC3
2     | Bug2    |        | 
2     | Bug2    | 1100   | TC2
2     | Bug2    | 1200   | TC3

编辑 2

完整请求

（TESTCYCL 在这篇文章中是 TEST）

SELECT disTINCT
B.BG_BUG_ID,B.BG_STATUS,B.BG_SUMMARY,B.BG_USER_TEMPLATE_13,B.BG_SEVERITY,B.BG_PRIORITY,B.BG_USER_TEMPLATE_10,B.BG_USER_TEMPLATE_08,R.REL_NAME,RC2.RCYC_NAME,B.BG_DETECTION_DATE,B.BG_RESPONSIBLE,B.BG_USER_TEMPLATE_03,B.BG_USER_03,B.BG_CLOSING_DATE,B.BG_USER_TEMPLATE_19,B.BG_USER_TEMPLATE_18,B.BG_DETECTED_BY,TC.TC_TESTCYCL_ID,TC.TC_TEST_ID,T.TS_NAME,TC.TC_STATUS,T.TS_USER_TEMPLATE_05,CF.CF_ITEM_NAME,C.CY_CYCLE,RC1.RCYC_NAME,TC.TC_USER_TEMPLATE_06,TC.TC_USER_TEMPLATE_05,TC.TC_USER_TEMPLATE_08
FROM TESTCYCL AS TC
INNER JOIN RELEASE_CYCLES AS RC1 ON TC.TC_ASSIGN_RCYC = RC1.RCYC_ID
INNER JOIN TEST AS T ON TC.TC_TEST_ID = T.TS_TEST_ID
INNER JOIN CYCLE AS C ON TC.TC_CYCLE_ID = C.CY_CYCLE_ID
INNER JOIN CYCL_FOLD AS CF ON C.CY_FOLDER_ID = CF.CF_ITEM_ID
FULL OUTER JOIN LINK AS L ON L.LN_ENTITY_ID = TC.TC_TESTCYCL_ID
FULL OUTER JOIN BUG AS B
INNER JOIN RELEASE_CYCLES AS RC2 ON B.BG_DETECTED_IN_RCYC = RC2.RCYC_ID
INNER JOIN RELEASES AS R ON B.BG_DETECTED_IN_REL = R.REL_ID
ON L.LN_BUG_ID = B.BG_BUG_ID

解决方法

您需要对连接重新排序。

将与一张表相关的所有 inner join 与该表放在一起。：

SELECT Bug.Id,Bug.Name,Test.Id,Test.Name
FROM Test
INNER JOIN Test_Detail ON Test_Detail.Test_id = Test.Id
INNER JOIN Release ON Release.Id = Test.Release_id
INNER JOIN Cycle ON Cycle.Release_id = Release.Id
FULL OUTER JOIN Link ON Link.Test_id = Test.Id
FULL OUTER JOIN Bug ON Link.Bug_id = Bug.Id

假设您也想将某些内容加入 Bug。为此，您需要嵌套 ON 条件：

SELECT Bug.Id,Test.Name
FROM Test
INNER JOIN Test_Detail ON Test_Detail.Test_id = Test.Id
INNER JOIN Release ON Release.Id = Test.Release_id
INNER JOIN Cycle ON Cycle.Release_id = Release.Id
FULL OUTER JOIN Link ON Link.Test_id = Test.Id
FULL OUTER JOIN Bug
      INNER JOIN Bug_Detail bd ON bd.Bug_id = Bug.id
  ON Link.Bug_id = Bug.Id

请注意，与流行观点相反，括号 () 在这里没有什么区别，尽管它们在视觉上很有用。重要的是 ON 子句的顺序：Bug_detail 内连接嵌套在 Bug

的全连接内 ,

你的内连接正在把完整的外连接变成别的东西。重新排列连接并使用括号：

SELECT Bug.Id,Test.Name
FROM (Test JOIN
      Test_Detail
      ON Test_Detail.Test_id = Test.Id JOIN
      Release
      ON Release.Id = Test.Release_id JOIN
      Cycle
      ON Cycle.Release_id = Release.Id
     ) FULL OUTER JOIN
     Link
     ON Link.Test_id = Test.Id FULL OUTER JOIN
     Bug
     ON Link.Bug_id = Bug.Id;