SQL :如何从长 xml 字符串中提取我想要的子字符串?

Question

我的SQL表中有一列包含一个XML值。 我必须读取这三个元素中的值：

<POSTN_ID>0000-0000H1-POS</POSTN_ID>
<ROLE_ID>0000-00002B-ROL</ROLE_ID>
<STATUS>1</STATUS>

在这种情况下，哪些 SQL 方法可以帮助我？

<parameter name="OrgHierarchyDatasets_diffgram">
<diffgr:diffgram xmlns:msdata="urn:schemas-microsoft-com:xml-msdata" xmlns:diffgr="urn:schemas-microsoft-com:xml-diffgram-v1">
<NewDataSet>
<PositionRoles diffgr:id="PositionRoles1" msdata:rowOrder="0" diffgr:hasChanges="inserted">
<POSTN_ID>0000-0000H1-POS</POSTN_ID>
<ROLE_ID>0000-00002B-ROL</ROLE_ID>
<STATUS>1</STATUS>
</PositionRoles>
</NewDataSet>
</diffgr:diffgram>
</parameter>
<parameter name="ExistNodeCheck" type="System.Boolean">True</parameter>

Answer 1

以下语法可以帮助您捕获定义的对象并将它们带到表模型中。

DECLARE @xml xml = '
<parameter name="OrgHierarchyDatasets_diffgram">
<diffgr:diffgram xmlns:msdata="urn:schemas-microsoft-com:xml-msdata" xmlns:diffgr="urn:schemas-microsoft-com:xml-diffgram-v1">
<NewDataSet>
<PositionRoles diffgr:id="PositionRoles1" msdata:rowOrder="0" diffgr:hasChanges="inserted">
<POSTN_ID>0000-0000H1-POS</POSTN_ID>
<ROLE_ID>0000-00002B-ROL</ROLE_ID>
<STATUS>1</STATUS>
</PositionRoles>
</NewDataSet>
</diffgr:diffgram>
</parameter>
<parameter name="ExistNodeCheck" type="System.Boolean">True</parameter>'
SELECT  
Tbl.Col.value('POSTN_ID[1]', 'nvarchar(50)') AS POSTN_ID,
Tbl.Col.value('ROLE_ID[1]', 'nvarchar(50)') AS ROLE_ID,
Tbl.Col.value('STATUS[1]', 'nvarchar(50)') AS STATUS
FROM   @xml.nodes('//PositionRoles') Tbl(Col) 

Answer 2

使用 xPath/xQuery：

select cast('<parameter name="OrgHierarchyDatasets_diffgram">
<diffgr:diffgram xmlns:msdata="urn:schemas-microsoft-com:xml-msdata" xmlns:diffgr="urn:schemas-microsoft-com:xml-diffgram-v1">
<NewDataSet>
<PositionRoles diffgr:id="PositionRoles1" msdata:rowOrder="0" diffgr:hasChanges="inserted">
<POSTN_ID>0000-0000H1-POS</POSTN_ID>
<ROLE_ID>0000-00002B-ROL</ROLE_ID>
<STATUS>1</STATUS>
</PositionRoles>
</NewDataSet>
</diffgr:diffgram>
</parameter>
<parameter name="ExistNodeCheck" type="System.Boolean">True</parameter>' as xml) as col
into #tab
go
select
col,
nod,
nod.query('./POSTN_ID/text()'),
nod.value('./ROLE_ID[1]','varchar(255)')
from (
select
*, col.query('//PositionRoles/node()') as nod
from #tab
) x