XSLT 删除 SOAP 信封但保留名称空间 [英] XSLT to remove SOAP envelope but leave namespaces

问题描述

我需要从soap消息中取出soap信封.为此，我想使用 XSLT，而不是 java.操作这种类型的xml会是更合适的解决方案.

I need to remove soap envelope from soap message. For that I want to use XSLT, not java. It would be more proper solution for operating such type of xml.

例如我有一条肥皂消息:

For e.g. I have a soap message:

<soapenv:Envelope xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/" 
                  xmlns:tar="namespace" 
                  xmlns:tar1="namespace">
    <soapenv:Header/>
    <soapenv:Body>
        <tar:RegisterUser>
            <tar1:Source>?</tar1:Source>
            <tar1:Profile>
                <tar1:EmailAddress>?</tar1:EmailAddress>

            </tar1:Profile>
        </tar:RegisterUser>
    </soapenv:Body>
</soapenv:Envelope>

我希望我的输出是这样的:

And I want my output to be something like this:

<tar:RegisterUser xmlns:tar="namespace" xmlns:tar1="namespace">
    <tar1:Source>?</tar1:Source>
    <tar1:Profile>
        <tar1:EmailAddress>?</tar1:EmailAddress>

    </tar1:Profile>
</tar:RegisterUser>

有人可以向我提供一些有关如何执行此操作的想法吗?

Can someone provide me with some ideas on how to do this?

推荐答案

这摆脱了 soapenv: 元素和命名空间声明.

This gets rid of the soapenv: elements and the namespace declaration.

<xsl:stylesheet 
  version="1.0"
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/"
>
  <xsl:output indent="yes" />

  <xsl:template match="@* | node()">
    <xsl:copy>
      <xsl:apply-templates select="@* | node()" />
    </xsl:copy>
  </xsl:template>

  <xsl:template match="soapenv:*">
    <xsl:apply-templates select="@* | node()" />
  </xsl:template>
</xsl:stylesheet>

结果:

<tar:RegisterUser xmlns:tar="namespace">
  <tar1:Source xmlns:tar1="namespace">?</tar1:Source>
  <tar1:Profile xmlns:tar1="namespace">
    <tar1:EmailAddress>?</tar1:EmailAddress>
  </tar1:Profile>
</tar:RegisterUser>

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