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我很难弄清楚如何将 XML PATH 添加到我的代码中以连接一些信息，更不用说理解 XML PATH 的工作原理了。在过去两天的大部分时间里，我都在努力解决这个问题，希望能得到一些帮助！

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())

样本数据：

| 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                                    |

预期输出：

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

如果有人可以帮助我，我将不胜感激。

先感谢您！！

杰米小号

4

3 回答 3

3

XML PATH 的工作原理

我将尝试使用此设置来解释：

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)

目标是为每个组创建一个以逗号分隔的名称列表的结果。

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。

此查询将创建一个 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>

上面的查询可以用在相关的子查询中，像这样为每个组创建 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，这可以通过在子查询中的每个项目名称中添加逗号来完成select ','+I.Name。它会在第一个值之前为您留下一个额外的逗号。您可以使用STUFF功能将其删除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
于 2012-08-29T07:47:07.897 回答
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小提琴

于 2012-08-30T05:28:20.663 回答
0

我相信我有不同的方式来实现你正在寻找的结果......

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，这会奏效。

于 2012-08-29T03:38:31.463 回答