我正在使用SQL Server 2014，我遇到了查询问题。我有三张桌子。 Report由ClothingObservation和HygieneObservation各10个组成。我这样做的方法是在两种类型的观察中分别引用10行ReportId Report，每个报告总共20次观察。我想选择一个报告的所有行。当我尝试这样做时，我得到100行。我的目标是获得10行或20行NULL值。这仅用于测试目的，因此Report仅包含1行，ClothingObservation和HygieneObservation各包含10行，所有行都引用现有报告的ReportId
我的表格，为清晰起见省略了详细信息：
CREATE TABLE HygieneObservation
(
HygieneObservationId int PRIMARY KEY IDENTITY NOT NULL,
...
ReportId int NOT NULL
)
CREATE TABLE ClothingObservation
(
ClothingObservationId int PRIMARY KEY IDENTITY NOT NULL,
...
ReportId int NOT NULL
)
CREATE TABLE Report
(
ReportId int PRIMARY KEY IDENTITY NOT NULL,
Period Date NOT NULL,
Reporter nvarchar(8) NOT NULL,
DepartmentId int NOT NULL
)
我的查询：
SELECT
Report.ReportId,
Report.Period,
Report.Reporter,
Report.DepartmentId,
ClothingObservation.ClothingObservationId,
HygieneObservation.HygieneObservationId
FROM Report
LEFT JOIN ClothingObservation ON
(ClothingObservation.ReportId = Report.ReportId)
LEFT JOIN HygieneObservation ON
(HygieneObservation.ReportId = Report.ReportId)
GROUP BY
Report.ReportId,
Period,
Reporter,
DepartmentId,
ClothingObservation.ClothingObservationId,
HygieneObservation.HygieneObservationId
这给了我100行，据我所知，因为ClothingObservation中的每一行都匹配HygieneObservation中的每一行。我认为使用GROUP BY会导致重复删除，但我显然做错了。任何提示？
编辑：这是我现在的数据（详情略去）。
Report:
ReportId Period Reporter DepartmentId
----------- ---------- -------- ------------
1 2016-05-01 username 1
ClothingObservation:
ClothingObservationId ... ReportId
--------------------- ... -----------
1 ... 1
2 ... 1
3 ... 1
4 ... 1
5 ... 1
6 ... 1
7 ... 1
8 ... 1
9 ... 1
10 ... 1
HygieneObservation:
HygieneObservationId ... ReportId
-------------------- ... -----------
3 ... 1
4 ... 1
5 ... 1
6 ... 1
7 ... 1
8 ... 1
9 ... 1
10 ... 1
12 ... 1
13 ... 1
编辑2：如果我运行这两个查询，我得到了我想要的输出（再次，从结果中省略了不相关的细节）：
SELECT * FROM Report
LEFT JOIN ClothingObservation ON
(ClothingObservation.ReportId = Report.ReportId)
SELECT * FROM Report
LEFT JOIN HygieneObservation ON
(HygieneObservation.ReportId = Report.ReportId)
ReportId Period Reporter DepartmentId ClothingObservationId ... ReportId
----------- ---------- -------- ------------ --------------------- ...- -----------
1 2016-05-01 username 1 1 ... 1
1 2016-05-01 username 1 2 ... 1
1 2016-05-01 username 1 3 ... 1
1 2016-05-01 username 1 4 ... 1
1 2016-05-01 username 1 5 ... 1
1 2016-05-01 username 1 6 ... 1
1 2016-05-01 username 1 7 ... 1
1 2016-05-01 username 1 8 ... 1
1 2016-05-01 username 1 9 ... 1
1 2016-05-01 username 1 10 ... 1
ReportId Period Reporter DepartmentId HygieneObservationId ... ReportId
----------- ---------- -------- ------------ -------------------- ... -----------
1 2016-05-01 username 1 3 ... 1
1 2016-05-01 username 1 4 ... 1
1 2016-05-01 username 1 5 ... 1
1 2016-05-01 username 1 6 ... 1
1 2016-05-01 username 1 7 ... 1
1 2016-05-01 username 1 8 ... 1
1 2016-05-01 username 1 9 ... 1
1 2016-05-01 username 1 10 ... 1
1 2016-05-01 username 1 12 ... 1
1 2016-05-01 username 1 13 ... 1
我的目标是通过一个查询获得此输出（或类似的内容）。
答案 0 :(得分：2)
您也可以尝试使用以下查询：
SELECT
ReportId = ISNULL(v1.ReportId, v2.ReportId),
Period = ISNULL(v1.Period, v2.Period),
Reporter = ISNULL(v1.Reporter, v2.Reporter),
DepartmentId = ISNULL(v1.DepartmentId, v2.DepartmentId),
v1.ClothingObservationId,
v2.HygieneObservationId
FROM
(
SELECT
RowNumber = ROW_NUMBER() OVER(Partition BY r.ReportId ORDER BY c.ClothingObservationId),
r.ReportId,
r.Period,
r.Reporter,
r.DepartmentId,
c.ClothingObservationId
FROM
Report r
LEFT JOIN ClothingObservation c ON c.ReportId = r.ReportId) v1
FULL JOIN
(
SELECT
RowNumber = ROW_NUMBER() OVER(Partition BY r.ReportId ORDER BY h.HygieneObservationId),
r.ReportId,
r.Period,
r.Reporter,
r.DepartmentId,
h.HygieneObservationId
FROM Report r
LEFT JOIN HygieneObservation h ON h.ReportId = r.ReportId) v2 ON v1.RowNumber = v2.RowNumber AND v1.ReportId = v2.ReportId
ORDER BY ReportId
答案 1 :(得分：1)
正在发生的事情是，将报告（1行）加入到ClothingObservation（10行）会产生10行（1 x 10），然后加入HygieneObservation（10行），这样就可以得到100.这种情况发生的原因是因为在初始连接之后，您有10行具有相同的ReportID，因此下一个连接将获取这10行中的每一行并连接到HygieneObservation中的10行。
SELECT
Report.ReportId,
Report.Period,
Report.Reporter,
Report.DepartmentId,
ClothingObservation.ClothingObservationId,
NULL AS HygieneObservationId
FROM Report
LEFT JOIN ClothingObservation ON
(ClothingObservation.ReportId = Report.ReportId)
UNION ALL
SELECT
Report.ReportId,
Report.Period,
Report.Reporter,
Report.DepartmentId,
NULL AS ClothingObservationId,
HygieneObservation.HygieneObservationId
FROM Report
LEFT JOIN HygieneObservation ON
(HygieneObservation.ReportId = Report.ReportId)
工作原理：
你基本上写了两个单独的查询：一个连接Report和ClothingObservation，另一个连接Report到HygieneObservation。然后，将两个查询与UNION ALL组合在一起。
这很复杂，因为它涉及我称之为“垂直合并”或“合并加入”。以下是查询（更新：我已对其进行了测试）。
SELECT
Report.ReportId,
Report.Period,
Report.Reporter,
Report.DepartmentId,
MergedObservations.ClothingObservationId,
MergedObservations.HygieneObservationId
FROM Report
LEFT JOIN
( SELECT COALESCE( ClothingObservation.ReportID, HygieneObservation.ReportID ) AS ReportID,
HygieneObservationID, ClothingObservationID -- Add appropriate columns
FROM
( SELECT ROW_NUMBER() OVER( PARTITION BY ReportID ORDER BY ClothingObservationID ) AS ResultID, ReportID, ClothingObservationID
FROM ClothingObservation ) AS ClothingObservation
FULL OUTER JOIN
( SELECT ROW_NUMBER() OVER( PARTITION BY ReportID ORDER BY HygieneObservationID ) AS ResultID, ReportID, HygieneObservationID
FROM HygieneObservation ) AS HygieneObservation
ON ClothingObservation.ReportID = HygieneObservation.ReportID
AND ClothingObservation.ResultID = HygieneObservation.ResultID
) AS MergedObservations
ON Report.ReportID = MergedObservations.ReportID
工作原理：
因为ClothingObservation和HygieneObservationId彼此不直接相关并且每个ReportID具有不同的行数，所以我使用ROW_NUMBER()函数来生成连接键。然后，我使用ReportID和ROW_NUMBER()函数的输出进行“合并连接”。
我已将您的示例数据转换为可用的表数据，以测试上述查询。
CREATE TABLE Report( ReportId INT, Period DATETIME, Reporter VARCHAR( 20 ), DepartmentId INT )
CREATE TABLE ClothingObservation( ClothingObservationID INT, ReportId INT )
CREATE TABLE HygieneObservation( HygieneObservationID INT, ReportId INT )
INSERT INTO Report
VALUES( 1, '2016-05-01', 'username', 1 )
INSERT INTO ClothingObservation
VALUES
( 1, 1 ), ( 2, 1 ), ( 3, 1 ), ( 4, 1 ), ( 5, 1 ), ( 6, 1 ), ( 7, 1 ), ( 8, 1 ), ( 9, 1 ), ( 10, 1 )
INSERT INTO HygieneObservation
VALUES
( 3, 1 ), ( 4, 1 ), ( 5, 1 ), ( 6, 1 ), ( 7, 1 ), ( 8, 1 ), ( 9, 1 ), ( 10, 1 ), ( 11, 1 ), ( 12, 1 ), ( 13, 1 )