0

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将其转置为：

<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>
4

2 回答 2

1

我怀疑使用 可能有一个非常好的方法PIVOT，但我不太了解它可以肯定地说。我在这里提供的工作。为了更好地格式化并提供评论，我已将其拆分为多个块：

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：

建立矩阵的大小：

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

切碎数据，用于ROW_NUMBER捕获索引（-1使其从零开始），并在索引上使用模数和整数除法来计算新的行号和列号：

,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FOR XML正确选择选项和查询结构，然后我们就完成了：

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')

样本输出：

<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>

适用于任何大小的“方形”数据。请注意缺乏健全性检查/错误处理。

于 2010-01-15T11:46:50.653 回答
0 
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
于 2010-04-14T18:14:50.350 回答