XSLT - 识别具有相同属性值模式的连续节点
我有这样的 xml，
<section>
<p id="ss_main">aa</p>
<p id="ss_chap">bb</p>
<p id="main">cc</p>
<p id="main">dd</p>
<p id="main">ee</p>
<p id="ss_main">ff</p>
<p id="main">gg</p>
<p id="main">hh</p>
<p id="main">ii</p>
<p id="main">jj</p>
<p id="ss_chap">xx</p>
<p id="ss_main">yy</p>
<p id="ss_chap">zz</p>
</section>
我的要求是放置名为 ss 开头.
what my requirement is place new nodes named <ss_start> and <ss_end> by covering existing nodes witch are start with ss.
所以输出应该是，
<section>
<ss_start/>
<p id="ss_main">aa</p>
<p id="ss_chap">bb</p>
<ss_end/>
<p id="main">cc</p>
<p id="main">dd</p>
<p id="main">ee</p>
<ss_start/>
<p id="ss_main">ff</p>
<ss_end/>
<p id="main">gg</p>
<p id="main">hh</p>
<p id="main">ii</p>
<p id="main">jj</p>
<ss_start/>
<p id="ss_chap">xx</p>
<p id="ss_main">yy</p>
<p id="ss_chap">zz</p>
<ss_end/>
</section>
我可以通过
<xsl:template match="p[@id='ss_main']">
<ss_start/>
<p id="ss_main"><xsl:apply-templates/></p>
<ss_end/>
</xsl:template>
但我正在努力寻找从 ss 开始的 id attr 的连续节点，并通过 <ss_start> 和 <ss_end>代码>.
but I'm struggling to find consecutive nodes that id attr starting from ss and cover them by <ss_start> and <ss_end>.
谁能建议我一种方法，我该怎么做?
Can anyone suggest me a method how can I do this?
XSLT 1.0 兄弟递归
在 XSLT 1.0 中，您可以按如下方式执行此操作，它使用一种称为 同级递归 的技术(尽管 同级遍历 可能是一个更好的术语).
XSLT 1.0 sibling recursion
In XSLT 1.0 you can do this as follows, which uses a technique called sibling recursion (though sibling traversal is probably a better term).
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
version="1.0">
<xsl:output indent="yes" />
<xsl:template match="node() | @*">
<xsl:copy>
<xsl:apply-templates select="node() | @*" />
</xsl:copy>
</xsl:template>
<xsl:template match="section">
<xsl:copy>
<xsl:apply-templates select="*[1]" />
</xsl:copy>
</xsl:template>
<xsl:template match="section/*[starts-with(@id, 'ss')]" priority="5">
<xsl:if test="self::*[not(preceding-sibling::*[1][starts-with(@id, 'ss')])]">
<ss_start />
</xsl:if>
<xsl:copy>
<xsl:apply-templates select="node() | @*" />
</xsl:copy>
<xsl:if test="self::*[not(following-sibling::*[1][starts-with(@id, 'ss')])]">
<ss_end />
</xsl:if>
<xsl:apply-templates select="following-sibling::*[1]" />
</xsl:template>
<xsl:template match="section/*">
<xsl:copy>
<xsl:apply-templates select="node() | @*" />
</xsl:copy>
<xsl:apply-templates select="following-sibling::*[1]" />
</xsl:template>
</xsl:stylesheet>
当针对您的输入运行时，将创建此输出:
Which, when run against your input, will create this output:
<?xml version="1.0" encoding="UTF-8"?>
<section>
<ss_start/>
<p id="ss_main">aa</p>
<p id="ss_chap">bb</p>
<ss_end/>
<p id="main">cc</p>
<p id="main">dd</p>
<p id="main">ee</p>
<ss_start/>
<p id="ss_main">ff</p>
<ss_end/>
<p id="main">gg</p>
<p id="main">hh</p>
<p id="main">ii</p>
<p id="main">jj</p>
<ss_start/>
<p id="ss_chap">xx</p>
<p id="ss_main">yy</p>
<p id="ss_chap">zz</p>
<ss_end/>
</section>
我现在看到您用 xslt-2.0 标记了您的问题，这意味着您可以使用分组.我将尝试使用 XSLT 2.0 中的示例进行更新.
I see now that you tagged your question with xslt-2.0, which means you can use grouping. I'll try to update with an example in XSLT 2.0.
在 XSLT 2.0 中，您可以使用布尔值 true/false 作为组相邻分组键，如下所示，这比上面的 XSLT 1.0 代码要短很多:
In XSLT 2.0, you can use a boolean true/false as the the group-adjacent grouping key as follows, which is quite a bit shorter than the XSLT 1.0 code above:
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
version="2.0">
<xsl:output indent="yes" />
<xsl:template match="node() | @*">
<xsl:copy>
<xsl:apply-templates select="node() | @*" />
</xsl:copy>
</xsl:template>
<xsl:template match="section">
<xsl:copy>
<xsl:for-each-group select="*" group-adjacent="starts-with(@id, 'ss')">
<xsl:if test="current-grouping-key()"><ss_start /></xsl:if>
<xsl:apply-templates select="current-group()" />
<xsl:if test="current-grouping-key()"><ss_end /></xsl:if>
</xsl:for-each-group>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>