Question

我有这个表结构。 YearPart，MonthPart和DatePart包含他们描述的内容...... EX：2011,1,19（分别）

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)
)

使用此样本数据：

INSERT INTO @agenda VALUES (2010, 12, 4, 'Test Record', '', '')
INSERT INTO @agenda VALUES (2011, 1, 3, 'Another Record', '', '')
INSERT INTO @agenda VALUES (2011, 1, 3, 'Fred Birthday', '', '')
INSERT INTO @agenda VALUES (2011, 1, 4, 'Work Day', '', '')
INSERT INTO @agenda VALUES (2011, 12, 6, '2nd Test Record', '', '')

我想要的是这样的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，我还没有能够让嵌套工作正常。我通常最终关闭分组（例如，我得到多个Year = 2011元素，当时应该只有一个）。

如果无法做到这一点，我总是可以在.NET网站上创建XML ...

Answer 1

可以做到。

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

SQL“For XML Path” - 嵌套结果

1 个答案: