我有XML
<getInquiryAboutListReturn xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<inquiryAbouts>
<inquiryAbout>
<code>Code</code>
<nameKk>Something</nameKk>
<nameRu>Something</nameRu>
<documents xsi:nil="true"/>
</inquiryAbout>
</inquiryAbouts>
</getInquiryAboutListReturn>
我想用XSLT处理它以复制所有XML
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
<xsl:output indent="yes" />
<xsl:template match="/">
<xsl:copy-of select="//getInquiryAboutListReturn/inquiryAbouts"/>
</xsl:template>
</xsl:stylesheet>
如何在没有<documents xsi:nil="true"/>
或没有xsi:nil =“true”的情况下复制所有XML?
所需的输出XML
<getInquiryAboutListReturn xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<inquiryAbouts>
<inquiryAbout>
<code>Code</code>
<nameKk>Something</nameKk>
<nameRu>Something</nameRu>
</inquiryAbout>
</inquiryAbouts>
</getInquiryAboutListReturn>
答案 0 :(得分:7)
这个简单的XSLT:
<?xml version="1.0"?>
<xsl:stylesheet
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
version="1.0">
<xsl:output omit-xml-declaration="no" indent="yes"/>
<xsl:strip-space elements="*"/>
<!-- TEMPLATE #1 -->
<xsl:template match="node()|@*">
<xsl:copy>
<xsl:apply-templates select="node()|@*"/>
</xsl:copy>
</xsl:template>
<!-- TEMPLATE #2 -->
<xsl:template match="*[@xsi:nil = 'true']" />
</xsl:stylesheet>
...应用于OP的源XML:
<?xml version="1.0"?>
<getInquiryAboutListReturn xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<inquiryAbouts>
<inquiryAbout>
<code>Code</code>
<nameKk>Something</nameKk>
<nameRu>Something</nameRu>
<documents xsi:nil="true"/>
</inquiryAbout>
</inquiryAbouts>
</getInquiryAboutListReturn>
...生成预期结果XML:
<?xml version="1.0"?>
<getInquiryAboutListReturn xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<inquiryAbouts>
<inquiryAbout>
<code>Code</code>
<nameKk>Something</nameKk>
<nameRu>Something</nameRu>
</inquiryAbout>
</inquiryAbouts>
</getInquiryAboutListReturn>
<强>说明强>