XSLT -在保留现有属性的同时向一组元素添加编号id属性

Question

对于像这样的文件(具有潜在的附加属性)：

<text>
...
<p>"<char key="Utterson">I</char> saw <char key="Hyde">Mr. Hyde</char> go in by the old dissecting-room door, <char key="Poole">Poole</char>," <char key="Utterson">he</char> said. "Is that right, when <char key="Jekyll">Dr. Jekyll</char> is from home?"</p>
<p>"Quite right, <char key="Utterson">Mr. Utterson, sir</char>," replied <char key="Poole">the servant</char>. "<char key="Hyde">Mr. Hyde</char> has a key."</p>
<p>"<char key="Jekyll">Your master</char> seems to repose a great deal of trust in <char key="Hyde">that young man</char>, <char key="Poole">Poole</char>," resumed <char key="Utterson">the other</char> musingly.</p>
<p>"Yes, <char key="Utterson">sir</char>, <char key="Jekyll">he</char> do indeed," said <char key="Poole">Poole</char>. "<char key="servants">We</char> have all orders to obey <char key="Hyde">him</char>."</p>
<p>"<char key="Utterson">I</char> do not think <char key="Utterson">I</char> ever met <char key="">Mr. Hyde</char>?" asked <char key="Utterson">Utterson</char>.</p>
...

​

我希望生成一个副本，为每个char添加一系列编号ids，例如：

<text>
...
<p>"<char key="Utterson" id="Utterson-1">I</char> saw <char key="Hyde" id="Hyde-1">Mr. Hyde</char> go in by the old dissecting-room door, <char key="Poole" id="Poole-1">Poole</char>," <char key="Utterson" id="Utterson-2">he</char> said. "Is that right, when <char key="Jekyll" id="Jekyll-1">Dr. Jekyll</char> is from home?"</p>
<p>"Quite right, <char key="Utterson" id="Utterson-3">Mr. Utterson, sir</char>," replied <char key="Poole" id="Poole-2">the servant</char>. "<char key="Hyde" id="Hyde-2">Mr. Hyde</char> has a key."</p>
<p>"<char key="Jekyll" id="Jekyll-2">Your master</char> seems to repose a great deal of trust in <char key="Hyde" id="Hyde-3">that young man</char>, <char key="Poole" id="Poole-3">Poole</char>," resumed <char key="Utterson" id="Utterson-4">the other</char> musingly.</p>
<p>"Yes, <char key="Utterson" id="Utterson-5">sir</char>, <char key="Jekyll" id="Jekyll3-">he</char> do indeed," said <char key="Poole" id="Poole-4">Poole</char>. "<char key="servants" id="servants-1">We</char> have all orders to obey <char key="Hyde" id="Hyde-4">him</char>."</p>
<p>"<char key="Utterson" id="Utterson-6">I</char> do not think <char key="Utterson" id="Utterson-7">I</char> ever met <char key="" id="-5">Mr. Hyde</char>?" asked <char key="Utterson" id="Utterson-8">Utterson</char>.</p>
...

​

而不丢失以前的任何属性，但是使用相反的键属性对每个id进行编号。这样的事情有可能吗？提前谢谢你。

Answer 1

您应该从这里的标识模板开始，在那里复制所有现有的节点和属性。

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

然后，您只需要一个模板来匹配char元素，在那里您可以添加新属性。您可以通过使用xsl:number来实现这一点。

<xsl:attribute name="id">
    <xsl:value-of select="@key" />
    <xsl:text>-</xsl:text>
    <xsl:variable name="key" select="@key" />
    <xsl:number level="any" count="char[@key = $key]"></xsl:number>
 </xsl:attribute>

试试这个XSLT

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
    <xsl:output method="xml" indent="yes" />

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

    <xsl:template match="char">
        <xsl:copy>
            <xsl:attribute name="id">
                <xsl:value-of select="@key" />
                <xsl:text>-</xsl:text>
                <xsl:variable name="key" select="@key" />
                <xsl:number level="any" count="char[@key = $key]"></xsl:number>
            </xsl:attribute>
            <xsl:apply-templates select="@*|node()"/>
        </xsl:copy>
    </xsl:template>
</xsl:stylesheet>