给定一个使用distinct-values()函数返回不同状态列表的变量,是否有办法在for-each循环中对变量进行标记化?
<States>
<State>AL</State>
<State>AL</State>
<State>NM</State>
</States>
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以下变量返回AL和NM,但我不能使用for-each迭代它.有没有解决的办法?
<xsl:variable name="FormStates" select="distinct-values(States/State)"/>
<xsl:for-each select="$FormStates">
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XSLT 2.0好的.
该distinct-values()函数返回一系列您应该能够迭代的值.结果是说"标记化".
fn:distinct-values('AL', 'AL', 'NL')返回序列('AL', 'NL').
如果输出变量,xsl:value-of它将返回字符串"AL NL",因为默认序列分隔符xsl:value-of是单个空格字符.这是您可以使用@separator属性更改的内容:
输入
<?xml version="1.0" encoding="UTF-8"?>
<States>
<State>AL</State>
<State>AL</State>
<State>NM</State>
</States>
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XSLT
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0">
<xsl:template match="/">
<xsl:variable name="FormStates" select="distinct-values(States/State)"/>
<xsl:comment>xsl:value-of</xsl:comment>
<xsl:value-of select="$FormStates" separator=":"/>
<xsl:comment>xsl:for-each</xsl:comment>
<xsl:for-each select="$FormStates">
<xsl:value-of select="."/>
<xsl:text>:</xsl:text>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
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产量
<?xml version="1.0" encoding="UTF-8"?>
<!--xsl:value-of-->
AL:NM
<!--xsl:for-each-->
AL:NM:
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