提取SQL Server中属性和元素不一致的嵌套XML数据
我有一个包含两列的现有表:唯一ID、xml格式的数据。后者具有嵌套结构,但未标准化(见下面的示例)。要将每个ID的所有xml信息提取为以下格式:提取SQL Server中属性和元素不一致的嵌套XML数据,sql,sql-server,xml,database,Sql,Sql Server,Xml,Database,我有一个包含两列的现有表:唯一ID、xml格式的数据。后者具有嵌套结构,但未标准化(见下面的示例)。要将每个ID的所有xml信息提取为以下格式: ID | Variable | Value 现有表格安排如下: ID | Data ---+------------------------------------------------------- 1 | <ID><Input_CCY style="Input"> USD </Input_CCY&g
ID | Variable | Value
现有表格安排如下:
ID | Data
---+-------------------------------------------------------
1 | <ID><Input_CCY style="Input"> USD </Input_CCY>.....
2 | <ID><Input_CCY style="Input"> GBP </Input_CCY>.....
我曾尝试在SQL中使用交叉应用
方法首先提取单个单元格信息,效果很好:
SELECT
T.[ID] as [ID],
a.b.value('fn:local-name(.)', 'NVARCHAR(100)') As [Node_1_Name],
a.b.value('.', 'NVARCHAR(100)') As [Node_1_Value] -- value,
c.b.value('.', 'NVARCHAR(100)') As [Node_1_Type] -- value
FROM
[dbo].[all] AS T
CROSS APPLY
T.[data].nodes('/risk/*') AS a(b)
OUTER APPLY
a.b.nodes('@*') AS c(b)
WHERE
a.b.value('fn:local-name(.)', 'VARCHAR(100)') != 'table' -- tables
此查询返回:
ID | Node_1_Name | Node_1_Value | Node_1_Type
----+-----------------+---------------+----------------
1 | Input_CCY | USD | Input
1 | Input_LastYear| 500 | Output
1 | Input_LastYear| 500 | =SUM(E65:E68)
2 | Input_CCY | GBP | Input
...
但是,当我添加更多级别的交叉应用
以深入到表的嵌套级别时:
SELECT
T.[ID] as [ID],
a.b.value('fn:local-name(.)', 'NVARCHAR(100)') As [Node_1_Name],
c.b.value('.', 'NVARCHAR(100)') As [Node_1_Attribute_Value],
g.b.value('fn:local-name(.)', 'NVARCHAR(100)') As [Node_2_Name],
h.b.value('.', 'NVARCHAR(100)') As [Node_2_Attribute_Value],
e.b.value('fn:local-name(.)', 'NVARCHAR(100)') As [Node_3_Name],
e.b.value('.', 'NVARCHAR(100)') As [Node_3_Element],
f.b.value('fn:local-name(.)', 'NVARCHAR(100)') As [Node_3_Attribute],
f.b.value('.', 'NVARCHAR(100)') As [Node_3_Attribute_Value]
FROM
[dbo].[all] AS T
CROSS APPLY
T.[data].nodes('/risk/*') AS a(b)
OUTER APPLY
a.b.nodes('@*') AS c(b)
CROSS APPLY
T.[data].nodes('/risk/table/*') g(b)
OUTER APPLY
g.b.nodes('@name') h(b)
CROSS APPLY
T.[data].nodes('/risk/table/Row/*') e(b)
OUTER APPLY
e.b.nodes('@name') f(b)
WHERE
a.b.value('fn:local-name(.)', 'VARCHAR(100)') = 'table'
AND c.b.value('fn:local-name(.)', 'NVARCHAR(100)') = 'name'
AND f.b.value('fn:local-name(.)', 'NVARCHAR(100)') = 'name'
结果表中似乎有大量重复项。例如,即使对于单个ID(即原始数据集中的1行),第一个表也会有30k+行(应该是8 x 24表=192行)。我怀疑我滥用了交叉应用程序
,但无法找到答案
这是我希望将表元素提取为的格式:
ID | Table Name | Row# | Row Name | Col# | Col Name | Value | Type (input/output)
或者我应该完全使用其他一些函数
我对单个单元格和表节点的编码进行了拆分,因为我认为在以后重新组合它们之前,先给定它们的结构,然后再为它们单独编码会更容易。如果有一种方法可以一炮打响,那就太好了 整个过程取决于你事先知道的内容 模拟您的问题的实体模型:
DECLARE @YourTable TABLE(ID INT IDENTITY, [Data] XML);
INSERT INTO @YourTable VALUES
(N'<risk>
<Input_CCY style="Input">USD</Input_CCY>
<Input_LastYear style="Output" formula="=SUM(E65:E68)">500</Input_LastYear>
<table name="info" header_row="1">
<Row name="" iRow="2">
<Col style="Input" name="Ref" iCol="1" >AAA</Col>
<Col style="Input" name="Location" iCol="2">London</Col>
...
</Row>
<Row name="" iRow="3">
<Col style="Input" name="Ref" iCol="1" >BBB</Col>
<Col style="Input" name="Location" iCol="2">Edinburgh</Col>
...
</Row>
...
</table>
<table name="summary" header_row="1">
<Row name="" iRow="2">
<Col style="Output" name="Status" iCol="1" >Amber</Col>
<Col style="Output" name="Referral_Bonus" iCol="2">No</Col>
...
</Row>
<Row name="" iRow="3">
<Col style="Normal" name="Status" iCol="1" >Green</Col>
<Col style="Normal" name="Referral_Bonus" iCol="2">YES</Col>
...
</Row>
...
</table>
</risk>');
--此查询将按已知名称选择列。--正如我在这里所说的,您将获得所有内容,但如果实际表中不存在所有列,则所有列都返回为NULL
SELECT t.ID
,t.[Data].value('(/risk/Input_CCY/@style)[1]','nvarchar(100)') AS Input_CCY_Style
,t.[Data].value('(/risk/Input_CCY/text())[1]','nvarchar(100)') AS Input_CCY_Content
,t.[Data].value('(/risk/Input_LastYear/@style)[1]','nvarchar(100)') AS Input_LastYear_Style
,t.[Data].value('(/risk/Input_LastYear/@formula)[1]','nvarchar(100)') AS Input_LastYear_Formula
,t.[Data].value('(/risk/Input_LastYear/text())[1]','nvarchar(100)') AS Input_LastYear_Content
,tbl.value('@name','nvarchar(250)') AS Table_Name
,rw.value('@iRow','int') AS Row_iRow
,rw.value('(Col[@name="Ref"]/@iCol)[1]','nvarchar(150)') AS Col_Ref_iCol
,rw.value('(Col[@name="Ref"]/@style)[1]','nvarchar(150)') AS Col_Ref_Style
,rw.value('(Col[@name="Ref"]/text())[1]','nvarchar(150)') AS Col_Ref_Content
,rw.value('(Col[@name="Location"]/@iCol)[1]','nvarchar(150)') AS Col_Location_iCol
,rw.value('(Col[@name="Location"]/@style)[1]','nvarchar(150)') AS Col_Location_Style
,rw.value('(Col[@name="Location"]/text())[1]','nvarchar(150)') AS Col_Location_Content
,rw.value('(Col[@name="Status"]/@iCol)[1]','nvarchar(150)') AS Col_Status_iCol
,rw.value('(Col[@name="Status"]/@style)[1]','nvarchar(150)') AS Col_Status_Style
,rw.value('(Col[@name="Status"]/text())[1]','nvarchar(150)') AS Col_Status_Content
,rw.value('(Col[@name="Referral_Bonus"]/@iCol)[1]','nvarchar(150)') AS Col_Referral_Bonus_iCol
,rw.value('(Col[@name="Referral_Bonus"]/@style)[1]','nvarchar(150)') AS Col_Referral_Bonus_Style
,rw.value('(Col[@name="Referral_Bonus"]/text())[1]','nvarchar(150)') AS Col_Referral_Bonus_Content
FROM @YourTable t
CROSS APPLY t.[Data].nodes('/risk/table') A(tbl)
CROSS APPLY tbl.nodes('Row') B(rw);
如果您事先知道所有的表和列,我建议您进行第二次评估,但每个表分别有一条语句。您提供的XML格式不正确。元素
有一个自动关闭标记,但有一个text()
节点和一个关闭标记:Amber
。我希望,您的原始数据在ìCol=“1”
之后没有/
…啊,那只是我试图发布数据的缩短版本并复制和粘贴。工作就像一种享受!作为后续:我最终使用了@Shungo建议的第一种方法,因为我事先不知道列名。但是,对于一个有20个风险的XML表,每个表大约有40个表(大约30行10列),脚本运行速度非常慢(约1小时)。有没有其他方法我可以用?在其他网站上也遇到过XMLIndex,但我不确定如何开始在我们的MS SQL server上实现它。…。@S.Poitier您提到的数字听起来并不多。。。一个小时很长。。。作为第一步,我将尝试分步骤地打破这一点,以找出在哪一点上消耗了这段时间。重要的是要知道:XML不是作为您看到的字符串存储的,而是作为层次结构表存储的。第一个解析非常昂贵,其余的解析速度非常快。您可以查看执行计划以查找详细信息。关于XML索引:在可以直接指定XPath的非常特殊的情况下,它们确实有很大帮助。在一个相当普遍的情况下,他们根本帮不上忙,因为你事先知道的还不够多。
SELECT t.ID
,t.[Data].value('(/risk/Input_CCY/@style)[1]','nvarchar(100)') AS Input_CCY_Style
,t.[Data].value('(/risk/Input_CCY/text())[1]','nvarchar(100)') AS Input_CCY_Content
,t.[Data].value('(/risk/Input_LastYear/@style)[1]','nvarchar(100)') AS Input_LastYear_Style
,t.[Data].value('(/risk/Input_LastYear/@formula)[1]','nvarchar(100)') AS Input_LastYear_Formula
,t.[Data].value('(/risk/Input_LastYear/text())[1]','nvarchar(100)') AS Input_LastYear_Content
,tbl.value('@name','nvarchar(250)') AS Table_Name
,rw.value('@iRow','int') AS Row_iRow
,cl.value('@name','nvarchar(250)') AS [Col_Name]
,cl.value('@iCol','nvarchar(250)') AS Col_iCol
,cl.value('text()[1]','nvarchar(150)') AS Col_Content
FROM @YourTable t
CROSS APPLY t.[Data].nodes('/risk/table') A(tbl)
CROSS APPLY tbl.nodes('Row') B(rw)
CROSS APPLY rw.nodes('Col') C(cl);
SELECT t.ID
,t.[Data].value('(/risk/Input_CCY/@style)[1]','nvarchar(100)') AS Input_CCY_Style
,t.[Data].value('(/risk/Input_CCY/text())[1]','nvarchar(100)') AS Input_CCY_Content
,t.[Data].value('(/risk/Input_LastYear/@style)[1]','nvarchar(100)') AS Input_LastYear_Style
,t.[Data].value('(/risk/Input_LastYear/@formula)[1]','nvarchar(100)') AS Input_LastYear_Formula
,t.[Data].value('(/risk/Input_LastYear/text())[1]','nvarchar(100)') AS Input_LastYear_Content
,tbl.value('@name','nvarchar(250)') AS Table_Name
,rw.value('@iRow','int') AS Row_iRow
,rw.value('(Col[@name="Ref"]/@iCol)[1]','nvarchar(150)') AS Col_Ref_iCol
,rw.value('(Col[@name="Ref"]/@style)[1]','nvarchar(150)') AS Col_Ref_Style
,rw.value('(Col[@name="Ref"]/text())[1]','nvarchar(150)') AS Col_Ref_Content
,rw.value('(Col[@name="Location"]/@iCol)[1]','nvarchar(150)') AS Col_Location_iCol
,rw.value('(Col[@name="Location"]/@style)[1]','nvarchar(150)') AS Col_Location_Style
,rw.value('(Col[@name="Location"]/text())[1]','nvarchar(150)') AS Col_Location_Content
,rw.value('(Col[@name="Status"]/@iCol)[1]','nvarchar(150)') AS Col_Status_iCol
,rw.value('(Col[@name="Status"]/@style)[1]','nvarchar(150)') AS Col_Status_Style
,rw.value('(Col[@name="Status"]/text())[1]','nvarchar(150)') AS Col_Status_Content
,rw.value('(Col[@name="Referral_Bonus"]/@iCol)[1]','nvarchar(150)') AS Col_Referral_Bonus_iCol
,rw.value('(Col[@name="Referral_Bonus"]/@style)[1]','nvarchar(150)') AS Col_Referral_Bonus_Style
,rw.value('(Col[@name="Referral_Bonus"]/text())[1]','nvarchar(150)') AS Col_Referral_Bonus_Content
FROM @YourTable t
CROSS APPLY t.[Data].nodes('/risk/table') A(tbl)
CROSS APPLY tbl.nodes('Row') B(rw);