如何在 SQL Server 中解析 SOAP XML 并显示为表格
我需要在 SQL Server 中解析一个 SOAP xml 并将其转换为表
I need to parser a SOAP xml in SQL Server and convert it to table
<soap:Envelope xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<soap:Body>
<ExecCommandResponse xmlns="http://tempuri.org/">
<ExecCommandResult>
<Result xmlns="">
<row>
<LOT>VERL5B3002PL</LOT>
<ID>115</ID>
<WH>710</WH>
<STPL>12</STPL>
</row>
<row>
<LOT>VERL68804EVN</LOT>
<ID>3716</ID>
<WH>771</WH>
<STPL>6</STPL>
</row>
</Result>
</ExecCommandResult>
</ExecCommandResponse>
</soap:Body>
</soap:Envelope>
我需要在 sql server 中解析一个 SOAP xml 并将其转换为表
I need to parser a SOAP xml in sql server and convert it to table
LOT | ID | WH | STPL
VERL68804EVN | 3716 | 771 | 6
推荐答案
使用最新函数来查询 XML.
Use the up-to-date functions to query XML.
您的 XML 在命名空间上看起来不是很干净.有两个默认命名空间,其中一个是空的...因此我会完全避免(屏蔽)它们.
Your XML is not very clean looking on the namespaces. There are two default namespaces, one of them empty... Therefore I would avoid (mask) them entirely.
DECLARE @xml XML=
'<soap:Envelope xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<soap:Body>
<ExecCommandResponse xmlns="http://tempuri.org/">
<ExecCommandResult>
<Result xmlns="">
<row>
<LOT>VERL5B3002PL</LOT>
<ID>115</ID>
<WH>710</WH>
<STPL>12</STPL>
</row>
<row>
<LOT>VERL68804EVN</LOT>
<ID>3716</ID>
<WH>771</WH>
<STPL>6</STPL>
</row>
</Result>
</ExecCommandResult>
</ExecCommandResponse>
</soap:Body>
</soap:Envelope>';
SELECT r.value('LOT[1]','varchar(max)') AS LOT
,r.value('ID[1]','int') AS ID
,r.value('WH[1]','int') AS WH
,r.value('STPL[1]','int') AS STPL
FROM @xml.nodes('/*:Envelope/*:Body/*:ExecCommandResponse/*:ExecCommandResult/*:Result/*:row') AS A(r)
--甚至更简单(甚至可以在没有 *:
的情况下工作):
--or even simpler (would even work without the *:
):
SELECT r.value('LOT[1]','varchar(max)') AS LOT
,r.value('ID[1]','int') AS ID
,r.value('WH[1]','int') AS WH
,r.value('STPL[1]','int') AS STPL
FROM @xml.nodes('//*:row') AS A(r)
总的来说,我会说:尽可能具体,因此建议第一个......
In general I'd say: Be as specific as possible, therefore rather suggest the first...
相关文章