命名空间,我有以下XML转换XML使用XSLT
<?xml version="1.0" encoding="UTF-8"?>
<typeNames xmlns="http://www.dsttechnologies.com/awd/rest/v1" xmlns:xs="http://www.w3.org/2001/XMLSchema">
<typeName recordType="case" href="awdServer/awd/services/v1/businessareas/SAMPLEBA/types/SAMPLECASE">SAMPLECASE</typeName>
<typeName recordType="folder" href="awdServer/awd/services/v1/businessareas/SAMPLEBA/types/SAMPLEFLD">SAMPLEFLD</typeName>
<typeName recordType="source" href="awdServer/awd/services/v1/businessareas/SAMPLEBA/types/SAMPLEST">SAMPLEST</typeName>
<typeName recordType="transaction" href="awdServer/awd/services/v1/businessareas/SAMPLEBA/types/SAMPLEWT">SAMPLEWT</typeName>
</typeNames>
我想通过使用XSLT如下转换上面的XML:
<response>
<results>
<source>
SAMPLEST
</source>
</results>
</response>
</xsl:template>
我只是想从输入源xml到输出xml。
我想用下面的XML,但无法获得所需的输出XML:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:v="http://www.dsttechnologies.com/awd/rest/v1" version="2.0" exclude-result-prefixes="v">
<xsl:output method="xml" version="1.0" omit-xml-declaration="yes" encoding="UTF-8" indent="yes" />
<xsl:strip-space elements="*" />
<!-- identity transform -->
<xsl:template match="@*|node()">
<xsl:copy>
<xsl:apply-templates select="@*|node()" />
</xsl:copy>
</xsl:template>
<xsl:template match="typeNames">
<response>
<results>
<source>
<xsl:value-of select="source" />
</source>
</results>
</response>
</xsl:template>
</xsl:stylesheet>
你已经声明了一个前缀,但你没有使用它。并且在输入XML –