Question

我正在努力弄清楚如何将XML PATH添加到我的代码中来连接一些信息，更不用说了解XML PATH的工作方式了。我花了最近两天的大部分时间来完成这项工作，并希望得到一些帮助!!

这是我正在使用的代码：

Select Top 100 Percent Agreements.AgrmntID, Agreements.Description As
  AgrmntDesc, Agreements.Status, AgreementSchedules.SchedDate, DateName(dw,
  AgreementSchedules.SchedDate), LaborCodeTypes.Description As LaborCode,
  Customers.CustName, Customers.CompanyName, JobSites.SiteName,
  AgreementSchedules.AgrmntSchedID
From Agreements Inner Join
  AgreementTypes On Agreements.AgrmntTypeID = AgreementTypes.AgrmntTypeID
  Inner Join
  AgreementSchedules On Agreements.AgrmntID = AgreementSchedules.AgrmntID
  Inner Join
  Customers On Agreements.CustID = Customers.CustID Inner Join
  JobSites On Agreements.CustSiteID = JobSites.CustSiteID Left Outer Join
  LaborCodeTypes On AgreementSchedules.RepairID = LaborCodeTypes.RepairID
Where Agreements.Status = 2 And Month(AgreementSchedules.SchedDate) =
  Month(GetDate())

示例数据：

| AgreementID | LaborCodeTypes.Description   | DateName(dw, AgreementSchedules.SchedDate)|
| - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -|
| 1           | Landscaping                  | Tuesday                                   |
| 1           | Landscaping                  | Friday                                    |
| 1           | Sweeping                     | Monday                                    |
| 1           | Sweeping                     | Wednesday                                 |
| 1           | Sweeping                     | Friday                                    |
| 2           | Landscaping                  | Monday                                    |

预期产出：

| AgreementID | LaborCode   | Days Of Week              |
| - - - - - - - - - - - - - - - - - - - - - - - - - - - |
| 1           | Landscaping | Tuesday, Friday           |
| 1           | Sweeping    | Monday, Wednesday, Friday |
| 2           | Landscaping | Monday                    |

如果有人能帮助我，我会非常感激。

提前谢谢!!

Jamie S

Answer 1

XML PATH如何工作

我将尝试使用此设置解释：

create table Grp
(
  GrpID int primary key,
  Name varchar(10)
)

create table Item
(
  ItemID int identity primary key,
  Name varchar(10),
  GrpID int references Grp(GrpID)
)

insert into Grp values
(1, 'G1'),
(2, 'G2')

insert into Item values
('A', 1),
('B', 1),
('C', 1),
('D', 2),
('E', 2)

目标是使用逗号分隔的每个组的名称列表创建结果。

GroupName  ItemNames
---------- ----------
G1         A,B,C
G2         D,E

FOR XML用于将查询结果转换为XML文档或XML片段。

此查询将创建XML片段。

select I.Name
from Item as I
for xml path(''), type

结果：

<Name>A</Name>
<Name>B</Name>
<Name>C</Name>
<Name>D</Name>
<Name>E</Name>

上面的查询可以在相关的子查询中使用，为这样的每个组创建XML片段。

select G.Name as GroupName,
       (
         select I.Name
         from Item as I
         where G.GrpID = I.GrpID
         for xml path(''), type
       ) as ItemNames
from Grp as G

结果：

GroupName  ItemNames
---------- --------------------------------------------
G1         <Name>A</Name><Name>B</Name><Name>C</Name>
G2         <Name>D</Name><Name>E</Name>

然后，您可以使用value()函数提取XML中的值。

select Name as GroupName,
       (
         select I.Name
         from Item as I
         where G.GrpID = I.GrpID
         for xml path(''), type
       ).value('.', 'varchar(max)') as ItemNames
from Grp as G

结果：

GroupName  ItemNames
---------- ----------
G1         ABC
G2         DE

要完成此操作，您需要添加逗号作为分隔符，并且可以通过向子查询select ','+I.Name中的每个项目名称添加逗号来完成。它将在第一个值之前为您留下额外的逗号。您可以使用STUFF功能删除STUFF(Value, 1, 1, '')。

最终查询：

select Name as GroupName,
       stuff((
         select ','+I.Name
         from Item as I
         where G.GrpID = I.GrpID
         for xml path(''), type
       ).value('.', 'varchar(max)'), 1, 1, '') as ItemNames
from Grp as G

Answer 2

;with C as
(
  select A.AgreementID,
         LCT.Description as LaborCode,
         Ags.ShedDate
  from Agreements as A
    inner join AgreementSchedules as AgS
      on A.AgreementID = AgS.AgreementID
    inner join LaborCodeTypes as LCT
      on AgS.RepairID = LCT.RepairID
  where A.[Status] = 2 and
        AgS.ShedDate >= dateadd(month, datediff(month, 0, getdate()), 0) and
        AgS.ShedDate < dateadd(month, 1+datediff(month, 0, getdate()), 0)
)
select C1.AgreementID,
       C1.LaborCode,
       stuff((select ', '+datename(weekday, C2.ShedDate)
              from C as C2
              where C1.AgreementID = C2.AgreementID and
                    C1.LaborCode = C2.LaborCode
              order by C2.ShedDate
              for xml path(''), type).value('.', 'varchar(max)'), 1, 2, '') as [Days Of Week]
from C as C1
group by C1.AgreementID, C1.LaborCode;

SQL Fiddle

Answer 3

我相信我有不同的方法来实现您正在寻找的结果......

SELECT * FROM (
    SELECT DISTINCT AgreementId, UPPER(Description), 1 AS IsLabor
    FROM Table
    WHERE ...
    UNION
    SELECT AgreementId, DateName AS Description, 0 AS IsLabor
    FROM Table
    WHERE ...
) x
ORDER BY AgreementId, IsLabor DESC, Description

这应该输出以下内容（UPPER只是为了强调劳动描述）：

| AgreementId | Description | IsLabor |
---------------------------------------
| 1           | LANDSCAPING | 1       |
| 1           | Friday      | 0       |
| 1           | Tuesday     | 0       |
| 2           | SWEEPING    | 1       |
| 2           | Friday      | 0       |
| 2           | Monday      | 0       |
| 2           | Wednesday   | 0       |

希望我理解你的问题，这将有效。

无法使FOR XML PATH正常工作

3 个答案: