SQL“For XML Path” – 嵌套结果

DECLARE @agenda AS TABLE    (
  PID INT IDENTITY(1,1) PRIMARY KEY,YearPart int,MonthPart int,DayPart int,lib_title nvarchar(200),[filename] nvarchar(255),meta_value nvarchar(2000)
)

使用此示例数据：

INSERT INTO @agenda VALUES (2010,12,4,'Test Record','','')
INSERT INTO @agenda VALUES (2011,3,'Another Record','Fred Birthday','Work Day',6,'2nd Test Record','')

我想要的是一个这样的XML输出：

<root>
  <Year Year="2010">
    <Month Month="12">
      <Day Day="4">
        <Item RecordName="Test Record" RecordID="1" />
      </Day>
    </Month>
  </Year>
  <Year Year="2011">
    <Month Month="1">
      <Day Day="3">
        <Item RecordName="Another Record" RecordID="2" />
        <Item RecordName="Geoffrey Birthday" RecordID="3" />
      </Day> 
      <Day Day="4">
        <Item RecordName="Work Day" RecordID="4" />
      </Day>
    </Month>
    <Month Month="12">
      <Day Day="6">
        <Item RecordName="2nd Test Record" RecordID="5" />
      </Day>
    </Month>
  </Year>
</root>

到目前为止,我还没有得到嵌套正常工作.我通常结束分组(例如,我得到多个Year = 2011元素,当只应该有一个).

如果这不能完成,我可以随时在.NET网站上创建XML

解决方法

select 
  a1.YearPart as '@Year',( select MonthPart as '@Month',(select DayPart as '@Day',(select
            lib_title as '@RecordName',PID as '@RecordID'
          from @agenda as a4
          where a4.DayPart = a3.DayPart and
                a4.MonthPart = a2.MonthPart and
                a4.YearPart = a1.YearPart
          for xml path('Item'),type         
         )
       from @agenda as a3
       where a3.YearPart = a1.YearPart and
             a3.MonthPart = a2.MonthPart
       group by a3.DayPart
       for xml path('Day'),type      
      )
    from @agenda as a2
    where a1.YearPart = a2.YearPart
    group by a2.MonthPart
    for xml path('Month'),type
  )  
from @agenda as a1
group by YearPart
for xml path('Year'),root

（编辑：李大同）

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