源XML:
<record>
<protein>AAA</protein>
<reference>1234</reference>
<reference>679</reference>
</record>
<record>
<protein>BBB</protein>
<reference>9876</reference>
</record>
<record>
<protein>CCC</protein>
<reference>9876</reference>
<reference>14846</reference>
<reference>982</reference>
</record>我刚开始使用XLST,一直无法找到解决问题的方法。我需要专门创建一个XML文件,其中记录中的数据根据列表中的值被拆分成一个或多个新记录。请注意,列表(参考)中的值的数量是高度可变的。
所需的XML:
<record>
<protein>AAA</protein>
<reference>1234</reference>
</record>
<record>
<protein>AAA</protein>
<reference>679</reference>
</record>
<record>
<protein>BBB</protein>
<reference>9876</reference>
</record>
<record>
<protein>CCC</protein>
<reference>9876</reference>
</record>
<record>
<protein>CCC</protein>
<reference>14846</reference>
</record>
<record>
<protein>CCC</protein>
<reference>982</reference>
</record>任何帮助都是非常感谢的。
发布于 2012-02-14 01:21:02
我必须调整您的示例XML输入文件。
<?xml version="1.0" encoding="UTF-8"?>
<records>
<record>
<protein>AAA</protein>
<reference>1234</reference>
<reference>679</reference>
</record>
<record>
<protein>BBB</protein>
<reference>9876</reference>
</record>
<record>
<protein>CCC</protein>
<reference>9876</reference>
<reference>14846</reference>
<reference>982</reference>
</record>
</records>然后,XSL本身可能如下所示:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:xs="http://www.w3.org/2001/XMLSchema" >
<xsl:output exclude-result-prefixes="xsl xs" indent="yes"/>
<xsl:template match="/records/record">
<xsl:for-each select="reference">
<xsl:element name="record">
<xsl:element name="protein">
<xsl:value-of select="../protein/text()"/>
</xsl:element>
<xsl:element name="reference">
<xsl:value-of select="text()"/>
</xsl:element>
</xsl:element>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>https://stackoverflow.com/questions/9264595
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