我有具有多节点结构的 XML，并希望通过使用 SQL Server 2012 搜索值“12345”来获取 XML 中的节点路径。
示例 XML:
<?xml version="1.0" encoding="UTF-8"?>
<Parent>
<DataExtensions>
<DataItem key="Key1" type="String">
<value>12345</value>
</DataItem>
<DataItem key="Key2" type="String">
<value>6789</value>
</DataItem>
</DataExtensions>
<TestKey1>12345</TestKey1>
<Answers>
<AnswersList>
<Entry key="Key2" type="Answer">
<value ID="Key1ID" Text="12345" />
</Entry>
<Entry key="Key3" type="Answer">
<value ID="Key3ID" Text="6789" />
</Entry>
</AnswersList>
</Answers>
<TestKey2>12345</TestKey2>
<TestKey3>67890</TestKey3>
<TestKey4>12345</TestKey3>
</Parent>
我可以通过为每个 select 语句编写来获取值，但我想获取整个 XML 中键值或值等于 12345 的节点路径。
表格中的期望输出为:
(/*/TestKey1/text())
(/*/TestKey2/text())
(/*/TestKey4/text())
(/*/DataExtensions/DataItem[@key=("Key1")]/value())
(/*/Answers/AnswersList/Entry[@key=("Key2")]/value/text())
请帮忙。
最佳答案
您可以尝试以下解决方案。它有很多强大的功能，包括 XML 可视化树表示。您的答案在输出的 [XPath] 列中。
T-SQL
DECLARE @xml XML =
N'<Parent>
<DataExtensions>
<DataItem key="Key1" type="String">
<value>12345</value>
</DataItem>
<DataItem key="Key2" type="String">
<value>6789</value>
</DataItem>
</DataExtensions>
<TestKey1>12345</TestKey1>
<Answers>
<AnswersList>
<Entry key="Key2" type="Answer">
<value ID="Key1ID" Text="6789"/>
</Entry>
<Entry key="Key3" type="Answer">
<value ID="Key3ID" Text="12345"/>
</Entry>
</AnswersList>
</Answers>
<TestKey2>12345</TestKey2>
<TestKey3>67890</TestKey3>
<TestKey4>12345</TestKey4>
</Parent>';
-- Solution # 2.
-- a thing of beauty.
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 @xml.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
WHERE Value = '12345';
T-SQL，短版
DECLARE @DocHandle INT;
EXEC sp_xml_preparedocument @DocHandle OUTPUT, @xml;
--SELECT * FROM OPENXML(@DocHandle,'/*');
;WITH rs AS
(
SELECT * FROM OPENXML(@DocHandle,'/*')
), cte AS
(
-- anchor
SELECT id
,ParentID
--, nodetype
, [text]
,CAST(id AS VARCHAR(100)) AS [Path]
,CAST('/' + rs.localname AS VARCHAR(1000))
+ N'['
+ CAST(ROW_NUMBER() OVER(ORDER BY (SELECT 1)) AS NVARCHAR)
+ N']' AS [XPath]
FROM rs
WHERE ParentID IS NULL
UNION ALL
--recursive member
SELECT t.id
,t.ParentID
--, nodetype = (SELECT nodetype FROM rs WHERE id = t.ParentID)
, t.[text]
, CAST(a.[Path] + ',' + CAST( t.ID AS VARCHAR(100)) AS VARCHAR(100)) AS [Path]
, CAST(a.[XPath] + '/' + IIF(t.nodetype = 2, '@', '')
+ t.localname AS VARCHAR(1000))
+ N'['
+ CAST(ROW_NUMBER() OVER(PARTITION BY t.localname ORDER BY (SELECT 1)) AS NVARCHAR)
+ N']' AS [XPath]
FROM rs AS t
INNER JOIN cte AS a ON t.ParentId = a.id
)
SELECT ID, ParentID, [Path]
, REPLACE([XPath],'[1]/#text[1]','') AS XPath, [text] AS [Value]
FROM cte
WHERE [text] IS NOT NULL
AND CAST([text] AS VARCHAR(30)) = '12345'
ORDER BY [Path];
EXEC sp_xml_removedocument @DocHandle;
输出
+----+----------+---------------------+--------------------------------------------------------------+-------+
| ID | ParentID | Path | XPath | Value |
+----+----------+---------------------+--------------------------------------------------------------+-------+
| 35 | 11 | 0,11,35 | /Parent[1]/TestKey1 | 12345 |
| 43 | 25 | 0,12,13,20,23,25,43 | /Parent[1]/Answers[1]/AnswersList[1]/Entry[2]/value[1]/@Text | 12345 |
| 31 | 6 | 0,2,3,6,31 | /Parent[1]/DataExtensions[1]/DataItem[1]/value | 12345 |
| 44 | 26 | 0,26,44 | /Parent[1]/TestKey2 | 12345 |
| 46 | 28 | 0,28,46 | /Parent[1]/TestKey4 | 12345 |
+----+----------+---------------------+--------------------------------------------------------------+-------+
关于sql-server - 如何按值获取XML中的节点路径？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/63608698/