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Home/ Questions/Q 547931
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T11:01:34+00:00 2026-05-13T11:01:34+00:00

Let’s say we have the following XML: <root> <row> <column>row 1 col 1</column> <column>row

  • 0

Let’s say we have the following 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>

How do I transpose this, using T-SQL XQuery to:

<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>
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-13T11:01:34+00:00Added an answer on May 13, 2026 at 11:01 am

    I suspect there might be a really nice approach using PIVOT, but I don’t know it well enough to be able to say for sure. What I offer here works. I have split it up into chunks for better formatting and to provide commentary:

    To start with let’s capture the example data

    -- 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
    

    Now the actual transposing code:

    Establish the size of the matrix:

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

    Shred the data, use ROW_NUMBER to capture the index (the -1 is to make it zero based), and use modulo and integer divide on the index to work out the new row and column numbers:

    ,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
    )
    

    With this CTE FlattenedInput available, all we now need to do is get the FOR XML options and query structure right and we’re done:

    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')
    

    Sample output:

    <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>
    

    Works on any size ‘square’ data. Note the lack of sanity checking / error handling.

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