代码之家 › 专栏 › 技术社区 › Paulo Santos

如何用T-SQL XQuery转换XML表?

xquery tsql

Paulo Santos · 技术社区 · 15 年前

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>

我该如何使用 T-SQL XQuery 到:

<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>

2 回复 | 直到 13 年前

AakashM 15 年前

我怀疑使用 PIVOT 但我不太清楚,不能肯定地说。我在这里提供的是有效的。为了更好地格式化和提供评论,我已经将它分成若干块:

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是:

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)
)

切碎数据,使用 ROW_NUMBER 捕获索引 -1 使其从零开始),并使用索引上的模和整数除来计算新的行和列编号:

,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
)

用这个CTE FlattenedInput 现在我们要做的就是 FOR XML 选项和查询结构正确,我们完成了:

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')

样品输出:

<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>

处理任何尺寸的“正方形”数据。注意缺乏健全性检查/错误处理。

animuson 13 年前

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