使用 FOR XML 控制 XML 元素嵌套

Control on XML elements nesting using FOR XML

我举例说明我的问题。事实上,我在尝试了很多方法后找到了解决方案,但我想问一下这个解决方案是否好,或者出于任何原因是否最好使用替代方法。 事实上,我需要控制元素的创建方式。

我首先制作了一个包含我需要的所有数据的视图,然后我通过多次加入该视图从该视图中进行选择。

我在这里使用局部变量而不是视图复制了 "complexity":

DECLARE @Employees table(  
    EmpID int NOT NULL,  
    Name nvarchar(50),  
    Surname nvarchar(50),  
    DateOfBirth date,
    DepartmentID int,
    AccessLevel int);
insert into  @Employees    values ('1', 'John','Doe','1980-01-31',100,5)
insert into  @Employees    values ('2', 'Mary','Rose','1971-02-27',102,3)
insert into  @Employees    values ('3', 'Luke','Perry','1995-12-01',104,1)

这是期望的结果(员工、部门和安全是不同的元素 - 我的问题是像本例一样创建员工部门和安全):

<employee Name="John" Surname="Doe" DateOfBirth="1980-01-31">
  <department DepartmentID="100">
    <security AccessLevel="5" />
  </department>
</employee>
<employee Name="Mary" Surname="Rose" DateOfBirth="1971-02-27">
  <department DepartmentID="102">
    <security AccessLevel="3" />
  </department>
</employee>
<employee Name="Luke" Surname="Perry" DateOfBirth="1995-12-01">
  <department DepartmentID="104">
    <security AccessLevel="1" />
  </department>
</employee>

正如我所说,我发现每个 xml 元素加入一次视图(这里是 table 变量)是一个解决方案:

-- declare @Employees table as above and then:
    select
      employee.Name,
      employee.Surname,
      employee.DateOfBirth,
      department.DepartmentID, 
      security.AccessLevel from @Employees employee
    join @Employees department on department.DepartmentID = employee.DepartmentID
    join @Employees security on security.AccessLevel = employee.AccessLevel
    for xml auto

这会产生所需的输出。

for xml auto 的多重连接技术是否有效?

在别名中使用@xml中生成attributes。更简单的方法来做到这一点

SELECT NAME         AS [@Name],
       Surname      AS [@Surname],
       DateOfBirth  AS [@DateOfBirth],
       DepartmentID AS [department/@DepartmentID],
       AccessLevel  AS [department/security/@AccessLevel]
FROM   @Employees
FOR xml path('employee') 

结果:

<employee Name="John" Surname="Doe" DateOfBirth="1980-01-31">
  <department DepartmentID="100">
    <security AccessLevel="5" />
  </department>
</employee>
<employee Name="Mary" Surname="Rose" DateOfBirth="1971-02-27">
  <department DepartmentID="102">
    <security AccessLevel="3" />
  </department>
</employee>
<employee Name="Luke" Surname="Perry" DateOfBirth="1995-12-01">
  <department DepartmentID="104">
    <security AccessLevel="1" />
  </department>
</employee>