将XML数据分解为SQL行时保留元素顺序

Ant*_*ull 5 sql xquery sql-server-2005

在将XML分解为SQL Server视图中的行时,如何返回元素序列?

样本输入:

<ol>
  <li>Smith</li>
  <li>Jones</li>
  <li>Brown</li>
</ol>
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期望的输出:

Sequence  Name
--------  -----------
    1     Smith
    2     Jones
    3     Brown
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现有观点:

CREATE VIEW OrderedList
AS
SELECT [Sequence] = CAST(NULL AS int)   -- TODO: Get ordinal position
       [Name] = b.b.value('.', 'nvarchar(max)')
FROM
(
    SELECT a = CAST('<ol><li>Smith</li><li>Jones</li><li>Brown</li></ol>' AS xml)
) a
CROSS APPLY a.a.nodes('/ol/li') b (b)
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Mik*_*son 11

您可以row_number()在xml节点上使用.

CREATE VIEW OrderedList
AS
SELECT [Sequence] = ROW_NUMBER() OVER(ORDER BY b.b),
       [Name] = b.b.value('.', 'nvarchar(max)')
FROM
(
    SELECT a = CAST('<ol><li>Smith</li><li>Jones</li><li>Brown</li></ol>' AS xml)
) a
CROSS APPLY a.a.nodes('/ol/li') b (b)
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参考:使用 Adam Machanic的DENSE_RANK唯一识别XML节点.