如何使用T-SQL XQuery转置XML表？

Question

假设我们有以下XML：

<root>
  <row>
    <column>row 1 col 1</column>
    <column>row 1 col 2</column>
    <column>row 1 col 3</column>
  </row>
  <row>
    <column>row 2 col 1</column>
    <column>row 2 col 2</column>
    <column>row 2 col 3</column>
  </row>
  <row>
    <column>row 3 col 1</column>
    <column>row 3 col 2</column>
    <column>row 3 col 3</column>
  </row>
</root>

如何使用 T-SQL XQuery 将其转换为：

<root>
    <column>
        <row>row 1 col 1</row>
        <row>row 2 col 1</row>
        <row>row 3 col 1</row>
    </column>
    <column>
        <row>row 1 col 2</row>
        <row>row 2 col 2</row>
        <row>row 3 col 2</row>
    </column>
    <column>
        <row>row 1 col 3</row>
        <row>row 2 col 3</row>
        <row>row 3 col 3</row>
    </column>
</root>

Answer 1

我怀疑使用PIVOT可能有一个非常好的方法，但我不太清楚它能够肯定地说出来。我在这提供的工作。我把它拆分成块以便更好地格式化并提供评论：

首先让我们捕获示例数据

-- Sample data
DECLARE @x3 xml

SET @x3 = '
<root>
  <row>
    <column>row 1 col 1</column>
    <column>row 1 col 2</column>
    <column>row 1 col 3</column>
  </row>
  <row>
    <column>row 2 col 1</column>
    <column>row 2 col 2</column>
    <column>row 2 col 3</column>
  </row>
  <row>
    <column>row 3 col 1</column>
    <column>row 3 col 2</column>
    <column>row 3 col 3</column>
  </row>
</root>
'

DECLARE @x xml
SET @x = @x3

-- @x is now our input

现在实际的转置代码：

确定矩阵的大小：

WITH Size(Size) AS
(
    SELECT CAST(SQRT(COUNT(*)) AS int) 
    FROM @x.nodes('/root/row/column') T(C)
)

粉碎数据，使用ROW_NUMBER来捕获索引（-1使其为零），并对索引使用模数和整数除法来计算 new < / em>行号和列号：

,Flattened(NewRow, NewCol, Value) AS
(
    SELECT
        -- i/@size as old_r, i % @size as old_c, 
        i % (SELECT TOP 1 Size FROM Size) AS NewRow, 
        i / (SELECT TOP 1 Size FROM Size) AS NewCol, 
        Value
    FROM (
        SELECT
            (ROW_NUMBER() OVER (ORDER BY C)) - 1 AS i, 
            C.value('.', 'nvarchar(100)') AS Value
        FROM @x.nodes('/root/row/column') T(C)
        ) ShreddedInput
)

有了这个CTE FlattenedInput，我们现在需要做的就是获得FOR XML选项和查询结构，我们已经完成了：

SELECT
    (
        SELECT Value 'column'
        FROM
            Flattened t_inner
        WHERE
            t_inner.NewRow = t_outer.NewRow
        FOR XML PATH(''), TYPE
    ) row
FROM
    Flattened t_outer
GROUP BY NewRow
FOR XML PATH(''), ROOT('root')

示例输出：

<root>
  <row>
    <column>row 1 col 1</column>
    <column>row 2 col 1</column>
    <column>row 3 col 1</column>
  </row>
  <row>
    <column>row 1 col 2</column>
    <column>row 2 col 2</column>
    <column>row 3 col 2</column>
  </row>
  <row>
    <column>row 1 col 3</column>
    <column>row 2 col 3</column>
    <column>row 3 col 3</column>
  </row>
</root>

适用于任何尺寸的'square'数据。请注意缺乏完整性检查/错误处理。

Answer 2

SET @x3 = ' 
<root> 
  <row> 
    <column>row 1 col 1</column> 
    <column>row 1 col 2</column> 
    <column>row 1 col 3</column> 
  </row> 
  <row> 
    <column>row 2 col 1</column> 
    <column>row 2 col 2</column> 
    <column>row 2 col 3</column> 
  </row> 
  <row> 
    <column>row 3 col 1</column> 
    <column>row 3 col 2</column> 
    <column>row 3 col 3</column> 
  </row> 
</root> 
' 

select @x3 = replace(@x3,'<row>','<rowtemp>')
select @x3 = replace(@x3,'<column>','<row>')
select @x3 = replace(@x3,'<rowtemp>','<column>')

select @x3