我想知道做了发言[@id=current()/@id]
如果ID是例如整数只能工作:为什么这个比较只适用于整型ID?
<elem id="1">
<elem id="1">
但是,如果我有:
<elem id="AAA">
<elem id="AAA">
它不工作。
我的XSL在那里失败:
<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema"
xmlns:fn="http://www.w3.org/2005/xpath-functions"
xmlns:tn="http://"
exclude-result-prefixes="xsl xs fn tn">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:strip-space elements="*" />
<xsl:template match="@*|node()">
<xsl:copy>
<xsl:apply-templates select="@*|node()" />
</xsl:copy>
</xsl:template>
<xsl:template match="genre/*">
<xsl:copy>
<xsl:apply-templates select="@*" />
<xsl:apply-templates select="
book[@id=current()/@id][@action='extend']
[not(preceding-sibling::book[@id=current()/@id][@action='borrow'])]" />
<xsl:for-each-group
select="book[@id=current()/@id][@action='borrow']
|
book[@id=current()/@id][@action='extend']
[preceding-sibling::book[@id=current()/@id][@action='borrow']]"
group-starting-with="book[@action='borrow']">
<xsl:for-each select="current-group()[1]">
<xsl:copy>
<xsl:apply-templates select="@*" />
<xsl:call-template name="merge-books-deeply">
<xsl:with-param name="books" select="current-group()" />
<xsl:with-param name="name-path" select="()" />
</xsl:call-template>
</xsl:copy>
</xsl:for-each>
</xsl:for-each-group>
<xsl:apply-templates select="
node()[ not(self::book[@id=current()/@id][@action=('borrow','extend')])]" />
</xsl:copy>
</xsl:template>
<xsl:function name="tn:children-on-path" as="element()*">
<xsl:param name="base" as="element()*" />
<xsl:param name="path" as="xs:string*" />
<xsl:choose>
<xsl:when test="fn:empty($base)">
<xsl:sequence select="()" />
</xsl:when>
<xsl:when test="fn:empty($path)">
<xsl:copy-of select="$base/*" />
</xsl:when>
<xsl:otherwise>
<xsl:sequence select="tn:children-on-path(
$base/*[name()=$path[1]],
$path[position() ne 1])" />
</xsl:otherwise>
</xsl:choose>
</xsl:function>
<xsl:template name="merge-books-deeply">
<xsl:param name="books" as="element()*" />
<xsl:param name="name-path" as="xs:string*" />
<xsl:for-each-group
select="tn:children-on-path($books,$name-path)"
group-by="name()">
<xsl:for-each select="current-group()[last()]" >
<xsl:copy>
<xsl:apply-templates select="@*" />
<xsl:call-template name="merge-books-deeply">
<xsl:with-param name="books" select="$books" />
<xsl:with-param name="name-path" select="$name-path,name()" />
</xsl:call-template>
<xsl:apply-templates select="text()" />
</xsl:copy>
</xsl:for-each>
</xsl:for-each-group>
</xsl:template>
</xsl:stylesheet>
输入样本,其中它的工作原理:
<root>
<library id="L1">
<genre id="a">
<shelf1 id="1">
<book id="1" action="borrow">
<attributes>
<user>John</user>
</attributes>
<other1>y</other1>
</book>
<book id="1" action="extend">
<attributes>
<user>Woo</user>
<length>3</length>
</attributes>
<other2>y</other2>
</book>
<book id="1" action="extend">
<attributes>
<length>2</length>
<condition>ok</condition>
</attributes>
<other3>y</other3>
</book>
<book id="2" action="extend">
<attributes>
<length>99</length>
<condition>not-ok</condition>
</attributes>
<other>y</other>
</book>
</shelf1>
<shelf1 id="b">...
</shelf1>
</genre>
<genre id="b">...
</genre>
</library>
</root>
如果我改变了书中的ID 1a
:
<root>
<library id="L1">
<genre id="a">
<shelf1 id="1">
<book id="1a" action="borrow">
<attributes>
<user>John</user>
</attributes>
<other1>y</other1>
</book>
<book id="1a" action="extend">
<attributes>
<user>Woo</user>
<length>3</length>
</attributes>
<other2>y</other2>
</book>
<book id="1a" action="extend">
<attributes>
<length>2</length>
<condition>ok</condition>
</attributes>
<other3>y</other3>
</book>
<book id="2" action="extend">
<attributes>
<length>99</length>
<condition>not-ok</condition>
</attributes>
<other>y</other>
</book>
</shelf1>
<shelf1 id="b">...
</shelf1>
</genre>
<genre id="b">...
</genre>
</library>
</root>
我的错误输出:(它根本不合并)
<root>
<library id="L1">
<genre id="a">
<shelf1 id="1">
<book id="a1" action="borrow">
<attributes>
<user>John</user>
</attributes>
<other1>y</other1>
</book>
<book id="a1" action="extend">
<attributes>
<user>Woo</user>
<length>3</length>
</attributes>
<other2>y</other2>
</book>
<book id="a1" action="extend">
<attributes>
<length>2</length>
<condition>ok</condition>
</attributes>
<other3>y</other3>
</book>
<book id="2" action="extend">
<attributes>
<length>99</length>
<condition>not-ok</condition>
</attributes>
<other>y</other>
</book>
</shelf1>
<shelf1 id="b">...
</shelf1>
</genre>
<genre id="b">...
</genre>
</library>
</root>
预期输出:
<root>
<library id="L1">
<genre id="a">
<shelf1 id="1">
<book id="1a" action="borrow">
<attributes>
<user>Woo</user>
<length>2</length>
<condition>ok</condition>
</attributes>
<other1>y</other1>
</book>
<book id="2" action="extend">
<attributes>
<length>99</length>
<condition>not-ok</condition>
</attributes>
<other>y</other>
</book>
</shelf1>
<shelf1 id="b">...
</shelf1>
</genre>
<genre id="b">...
</genre>
</library>
</root>
对于动作的每一个节点=借随后用动作的一个或多个节点=延伸
合并一起与操作的节点=借。
将子属性合并为一个子属性,使它具有来自具有最新值的兄弟的所有唯一属性。
保持不变的
其他孩子合并只能与具有相同ID的节点发生。
谢谢。 Regards, John
它应该工作,你有什么证据证明它不是? – 2012-07-16 22:17:01
@MichaelKay请看看添加的例子,我是否缺少(非常感谢) – John 2012-07-17 00:39:01
@John,缺少的是(a)描述了结果是什么,(b)描述了你期望得到的结果。“不起作用”太模糊了。此外,如果您的示例数据可用,它将有所帮助; as它是,它不是一个格式良好的XML文档。) – LarsH 2012-07-17 02:01:19