Question

问题：

我正在尝试返回一个包含所有连接在一起的父项名称的列的联系人列表。请不要打扰告诉我这违反了第1范式，我已经知道了。我一直试图让这个工作很长时间，这可能是一个愚蠢的错误，但我现在需要一些帮助。

查询：

SELECT  C.ID, C.SequenceNumber, C.ContactType, C.GUID, C.ContactCategory, MAP.Parents, C.LastName, C.FirstName, C.CompanyName, C.Title, C.Email, C.Phone, C.MobilePhone, C.Fax, C.Comments, C.StandardConfirmation, C.Active FROM
(
SELECT ContactTypeMapping.Contact_GUID, STUFF( ( SELECT ','+ [NAME]
                        FROM ContactParents a
                        WHERE a.GUID =b.GUID
                        FOR XML PATH('')),1 ,1, '') Parents

FROM ContactParents b, ContactTypeMapping
WHERE ContactTypeMapping.Parent_GUID=b.GUID 
) MAP
INNER JOIN
(
SELECT Contact.ID, Contact.GUID, Contact.SequenceNumber, 
       Contact.ContactType, Contact.ContactCategory, Contact.LastName, Contact.FirstName, 
       Contact.CompanyName, Contact.Title, Contact.Email, Contact.Phone, 
       Contact.MobilePhone, Contact.Fax, Contact.Comments, Contact.StandardConfirmation, Contact.Active
      FROM Contact
       )C
     ON (MAP.Contact_GUID=C.GUID)

目前的结果：

ID   SequenceNumber  ContactType GUID                                    ContactCategory Parents         LastName    FirstName   CompanyName     Title       Email                       Phone           MobilePhone Fax             Comments    StandardConfirmation    Active
15   4              2           95A566D0-DB83-4853-9CB7-E6CF3B1FF814    0               AParent         Beard       Kirk        NULL            Business    Kirk_Beard@someplace.com    913-906-3333    NULL        (913) 906-3434  NULL        0                       1
15   4              2           95A566D0-DB83-4853-9CB7-E6CF3B1FF814    0               AnotherParent   Beard       Kirk        NULL            Business    Kirk_Beard@someplace.com    913-906-3333    NULL        (913) 906-3434  NULL        0                       1

期望的结果：


15  4   2   95A566D0-DB83-4853-9CB7-E6CF3B1FF814    0   AParent,AnotherParent Beard   Kirk    NULL    Business    Kirk_Beard@someplace.com    913-906-3333    NULL    (913) 906-3434  NULL    0   1

数据：

ContactParents表：


GUID                                    NAME            TYPE
C40A6F7E-F760-48D6-8BAF-E55EC7DC900D    AParent         Place
A651A0A3-5A50-45F1-AB4B-2B7FDCE9734C    AnotherParent   Place

ContactTypeMapping表：


Contact_GUID                            Parent_GUID                        ParentTable
95A566D0-DB83-4853-9CB7-E6CF3B1FF814    C40A6F7E-F760-48D6-8BAF-E55EC7DC900D   Place
95A566D0-DB83-4853-9CB7-E6CF3B1FF814    A651A0A3-5A50-45F1-AB4B-2B7FDCE9734C   Place

联系表：


ID  GUID                                   SequenceNumber   ContactType LastName    FirstName   Title       Email                       Phone          MobilePhone  Fax            Comments Active  ContactCategory CompanyName StandardConfirmation
15  95A566D0-DB83-4853-9CB7-E6CF3B1FF814    4               2           Beard       Kirk        Business    Kirk_Beard@someplace.com    913-906-3333    NULL        (913) 906-3434  NULL     1        0              NULL        0

Answer 1

;WITH cp AS -- contact parents - initial join
(
    SELECT cp.NAME, ctm.Parent_GUID
      FROM ContactParents AS cp
      INNER JOIN ContactTypeMapping AS ctm
      ON cp.GUID = ctm.Contact_GUID
),
cm AS -- contact mapping with concatenated values
(
    SELECT Parent_GUID, Parents = (
            SELECT STUFF ((SELECT ','+ [NAME] FROM cp AS cp2 
              WHERE cp2.Parent_GUID = cp.Parent_GUID 
              FOR XML PATH('')),1 ,1, '')
          )
      FROM cp GROUP BY Parent_GUID
)
SELECT c.[GUID], cm.Parents --, other columns from c 
  FROM Contact AS c
  INNER JOIN cm
  ON c.[GUID] = cm.Parent_GUID;

为了演示我如何验证此查询返回正确的结果，这里是我在本地创建的表变量，我如何填充它们，以及引用表变量的稍微不同的查询：

DECLARE @ContactParents TABLE
(
    [GUID] UNIQUEIDENTIFIER,
    NAME VARCHAR(32),
    [TYPE] VARCHAR(32)
);

INSERT @ContactParents VALUES
('C40A6F7E-F760-48D6-8BAF-E55EC7DC900D','AParent','Place'),
('A651A0A3-5A50-45F1-AB4B-2B7FDCE9734C','AnotherParent','Place');

DECLARE @ContactTypeMapping TABLE
(
    Contact_GUID UNIQUEIDENTIFIER,
    Parent_GUID UNIQUEIDENTIFIER,
    ParentTable VARCHAR(32)
);

INSERT @ContactTypeMapping VALUES
('A651A0A3-5A50-45F1-AB4B-2B7FDCE9734C','95A566D0-DB83-4853-9CB7-E6CF3B1FF814','Place'),
('C40A6F7E-F760-48D6-8BAF-E55EC7DC900D','95A566D0-DB83-4853-9CB7-E6CF3B1FF814','Place');

DECLARE @Contact TABLE
(
    ID INT,
    [GUID] UNIQUEIDENTIFIER,
    LastName VARCHAR(32),
    FirstName VARCHAR(32)
    --, other columns...
);

INSERT @Contact VALUES
(15, '95A566D0-DB83-4853-9CB7-E6CF3B1FF814', 'Beard', 'Kirk');

;WITH cp AS
(
    SELECT cp.NAME, ctm.Parent_GUID
      FROM @ContactParents AS cp
      INNER JOIN @ContactTypeMapping AS ctm
      ON cp.GUID = ctm.Contact_GUID
),
cm AS
(
    SELECT Parent_GUID, Parents = (
      SELECT STUFF ((SELECT ','+ [NAME] FROM cp AS cp2 
        WHERE cp2.Parent_GUID = cp.Parent_GUID 
        FOR XML PATH('')),1 ,1, '')
      )
    FROM cp
    GROUP BY Parent_GUID
)
SELECT c.[GUID], cm.Parents --, other columns from c
  FROM @Contact AS c
  INNER JOIN cm
  ON c.[GUID] = cm.Parent_GUID;

结果

GUID                                    Parents
95A566D0-DB83-4853-9CB7-E6CF3B1FF814    AParent,AnotherParent

试图连接多行..意外结果

1 个答案: