Вы можете попробовать следующее, чтобы преобразовать ЛЮБОЙ файл XML в таблицу SQL. Он использует формат xml, чтобы определить, является ли это новой записью или нет. Он также разделяет атрибуты как новую запись, но помечает их. Вы можете легко восстановить xml
ИСПОЛЬЗОВАТЬ:
select * from Utility.FlattenXml('your xml’)
ПРОЦЕДУРА:
USE [YOUR_DB]
GO
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
CREATE FUNCTION [Utility].[FlattenXml](@xmlDoc 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 @xmlDoc.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 into [your table]
ROW_NUMBER() OVER(ORDER BY Sort, ID) AS ID,
ParentName, ParentPosition,Depth, NodeName, Position,
NodeType, FullPath, XPath, TreeView, Value, XMLData
FROM CTE2
person
Ray
schedule
17.08.2020
FROM
(возможно, используя методnodes
), чтобы вернуть несколько элементовTransaction_ID
перед извлечением текста элемента с помощью методаvalue
. Добавьте полный запрос к своему вопросу, если вам нужна помощь в этом. - person Dan Guzman   schedule 25.06.2020