Question

我在SQL Server 2k8中有一个带有XML列的表。以下SQL检索一些XML：

SELECT TOP 1 my_xml_column FROM my_table

让我们说它返回以下XML

<a>
  <b />
  <c>
    <d />
    <d />
    <d />
  </c>
</a>

我想得到的是

/a
/a/b
/a/c
/a/c/d
/a/e

换句话说，我怎样才能让SQL Server告诉我XML的结构？

我可以执行以下操作来获取各个元素的所有名称：

SELECT  C1.query('fn:local-name(.)')
FROM    my_table
CROSS APPLY my_xml_column.nodes('//*') AS T ( C1 )

也许如果有一个等效于“local-name（）”的东西会返回该元素的整个路径呢？

Answer 1

您可以使用XQuery和递归CTE（无OPENXML）干净利落地完成此任务：

DECLARE @xml xml
SET @xml = '<a><b /><c><d /><d /><d /></c></a>';

WITH Xml_CTE AS
(
    SELECT
        CAST('/' + node.value('fn:local-name(.)',
            'varchar(100)') AS varchar(100)) AS name,
        node.query('*') AS children
    FROM @xml.nodes('/*') AS roots(node)

    UNION ALL

    SELECT
        CAST(x.name + '/' + 
            node.value('fn:local-name(.)', 'varchar(100)') AS varchar(100)),
        node.query('*') AS children
    FROM Xml_CTE x
    CROSS APPLY x.children.nodes('*') AS child(node)
)
SELECT DISTINCT name
FROM Xml_CTE
OPTION (MAXRECURSION 1000)

它并没有真正做很多XQuery魔术，但至少它是全内联的，不需要任何存储过程，特殊权限等。

Answer 2

你的UDF .....

SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
CREATE FUNCTION [dbo].[XMLTable](@x XML)  
RETURNS TABLE 
AS RETURN 
WITH cte AS (  
SELECT 
        1 AS lvl,  
        x.value('local-name(.)','NVARCHAR(MAX)') AS Name,  
        CAST(NULL AS NVARCHAR(MAX)) AS ParentName, 
        CAST(1 AS INT) AS ParentPosition, 
        CAST(N'Element' AS NVARCHAR(20)) AS NodeType,  
        x.value('local-name(.)','NVARCHAR(MAX)') AS FullPath,  
        x.value('local-name(.)','NVARCHAR(MAX)')  
        + N'[' 
        + CAST(ROW_NUMBER() OVER(ORDER BY (SELECT 1)) AS NVARCHAR)  
        + N']' AS XPath,  
        ROW_NUMBER() OVER(ORDER BY (SELECT 1)) AS Position, 
        x.value('local-name(.)','NVARCHAR(MAX)') AS Tree,  
        x.value('text()[1]','NVARCHAR(MAX)') AS Value,  
        x.query('.') AS this,         
        x.query('*') AS t,  
        CAST(CAST(1 AS VARBINARY(4)) AS VARBINARY(MAX)) AS Sort,  
        CAST(1 AS INT) AS ID  
FROM @x.nodes('/*') a(x)  
UNION ALL 
SELECT 
        p.lvl + 1 AS lvl,  
        c.value('local-name(.)','NVARCHAR(MAX)') AS Name,  
        CAST(p.Name AS NVARCHAR(MAX)) AS ParentName, 
    CAST(p.Position AS INT) AS ParentPosition, 
        CAST(N'Element' AS NVARCHAR(20)) AS NodeType,  
        CAST(p.FullPath + N'/' + c.value('local-name(.)','NVARCHAR(MAX)') AS NVARCHAR(MAX)) AS FullPath,  
        CAST(p.XPath + N'/'+ c.value('local-name(.)','NVARCHAR(MAX)')+ N'['+ CAST(ROW_NUMBER() OVER(PARTITION BY c.value('local-name(.)','NVARCHAR(MAX)') 
        ORDER BY (SELECT 1)) AS NVARCHAR)+ N']' AS NVARCHAR(MAX)) AS XPath,  
        ROW_NUMBER() OVER(PARTITION BY c.value('local-name(.)','NVARCHAR(MAX)')
        ORDER BY (SELECT 1)) AS Position, 
        CAST( SPACE(2 * p.lvl - 1) + N'|' + REPLICATE(N'-', 1) + c.value('local-name(.)','NVARCHAR(MAX)') AS NVARCHAR(MAX)) AS Tree,  
        CAST( c.value('text()[1]','NVARCHAR(MAX)') AS NVARCHAR(MAX) ) AS Value, c.query('.') AS this,  
        c.query('*') AS t,  
        CAST(p.Sort + CAST( (lvl + 1) * 1024 + (ROW_NUMBER() OVER(ORDER BY (SELECT 1)) * 2) AS VARBINARY(4)) AS VARBINARY(MAX) ) AS Sort,  
        CAST((lvl + 1) * 1024 + (ROW_NUMBER() OVER(ORDER BY (SELECT 1)) * 2) AS INT)  
FROM cte p  
CROSS APPLY p.t.nodes('*') b(c)), cte2 AS (  
                                            SELECT 
                                            lvl AS Depth,  
                                            Name AS NodeName,  
                                            ParentName, 
                                            ParentPosition, 
                                            NodeType,  
                                            FullPath,  
                                            XPath,  
                                            Position, 
                                            Tree AS TreeView,  
                                            Value,  
                                            this AS XMLData,  
                                            Sort, ID  
                                            FROM cte  
UNION ALL 
SELECT 
        p.lvl,  
        x.value('local-name(.)','NVARCHAR(MAX)'),  
        p.Name, 
        p.Position, 
        CAST(N'Attribute' AS NVARCHAR(20)),  
        p.FullPath + N'/@' + x.value('local-name(.)','NVARCHAR(MAX)'),  
        p.XPath + N'/@' + x.value('local-name(.)','NVARCHAR(MAX)'),  
        1, 
        SPACE(2 * p.lvl - 1) + N'|' + REPLICATE('-', 1)  
        + N'@' + x.value('local-name(.)','NVARCHAR(MAX)'),  
        x.value('.','NVARCHAR(MAX)'),  
        NULL,  
        p.Sort,  
        p.ID + 1  
FROM cte p  
CROSS APPLY this.nodes('/*/@*') a(x)  
)  
SELECT 
        ROW_NUMBER() OVER(ORDER BY Sort, ID) AS ID,  
        ParentName, ParentPosition,Depth, NodeName, Position,   
        NodeType, FullPath, XPath, TreeView, Value, XMLData 
FROM cte2

Answer 3

我怀疑SQL Server的XQuery实现不适合这项任务，但这是另一种方法（受this启发，根据需要进行调整）：

DECLARE @idoc INT, @xml XML
SET @xml = (SELECT TOP 1 my_xml_column FROM my_table)
EXEC sp_xml_preparedocument @idoc OUTPUT, @xml;

WITH
    E AS (SELECT * FROM OPENXML(@idoc,'/',3)),
    P AS
    (
    -- anchor member
    SELECT id, parentid, localname AS [Path]
    FROM E WHERE parentid IS NULL
    UNION ALL
    -- recursive member
    SELECT E.id, E.parentid, P.[Path] + '/' + localname AS [Path]
    FROM P INNER JOIN E ON E.parentid = P.id
    )
SELECT [Path] FROM P

EXEC sp_xml_removedocument @idoc

如何从SQL Server中的XML值获取元素名称列表

3 个答案: