Html 使用XPath在两个节点之间选择同级
如何选择id为header_completed的表和header_completed后的第一个表之间的所有表，即中间对齐的表？以下是我从中选择的html：Html 使用XPath在两个节点之间选择同级,html,xpath,Html,Xpath,如何选择id为header_completed的表和header_completed后的第一个表之间的所有表，即中间对齐的表？以下是我从中选择的html： <table border="0" cellpadding="0" cellspacing="0" width="920" align="center"></table> <table border="0" cellpadding="0" cellspacing="0" width="920"></t
<table border="0" cellpadding="0" cellspacing="0" width="920" align="center"></table>
<table border="0" cellpadding="0" cellspacing="0" width="920"></table>
<table border="0" cellpadding="0" cellspacing="0" width="920"></table>
<table border="0" cellpadding="0" cellspacing="0" width="920" align="center" class="header_completed"></table>
<table border="0" cellpadding="0" cellspacing="0" width="920"></table> <--
<table border="0" cellpadding="0" cellspacing="0" width="920"></table> <--
<table border="0" cellpadding="0" cellspacing="0" width="920"></table> <-- these 5
<table border="0" cellpadding="0" cellspacing="0" width="920"></table> <--
<table border="0" cellpadding="0" cellspacing="0" width="920"></table> <--
<table border="0" cellpadding="0" cellspacing="0" width="920" align="center"></table>
<table border="0" cellpadding="0" cellspacing="0" width="920"></table>
<table border="0" cellpadding="0" cellspacing="0" width="920"></table>
<table border="0" cellpadding="0" cellspacing="0" width="920" align="center"></table>
使用节点集相交的Kayesian方法：
<t>
<table border="0" cellpadding="0" cellspacing="0" width="920" align="center"></table>
<table border="0" cellpadding="0" cellspacing="0" width="920"></table>
<table border="0" cellpadding="0" cellspacing="0" width="920"></table>
<table border="0" cellpadding="0" cellspacing="0" width="920" align="center" class="header_completed"></table>
<table border="0" cellpadding="0" cellspacing="1" width="920"></table>
<table border="0" cellpadding="0" cellspacing="2" width="920"></table>
<table border="0" cellpadding="0" cellspacing="3" width="920"></table>
<table border="0" cellpadding="0" cellspacing="4" width="920"></table>
<table border="0" cellpadding="0" cellspacing="5" width="920"></table>
<table border="0" cellpadding="0" cellspacing="0" width="920" align="center"></table>
<table border="0" cellpadding="0" cellspacing="0" width="920"></table>
<table border="0" cellpadding="0" cellspacing="0" width="920"></table>
<table border="0" cellpadding="0" cellspacing="0" width="920" align="center"></table>
</t>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:variable name="ns1" select=
"/*/*[@class='header_completed'][1]
/following-sibling::*
"/>
<xsl:variable name="ns2" select=
"/*/*[@class='header_completed'][1]
/following-sibling::*[@align='center'][1]
/preceding-sibling::*
"/>
<xsl:template match="/">
<xsl:copy-of select=
"$ns1[count(.| $ns2)=count($ns2)]
"/>
<DELIMITER/>
<xsl:copy-of select=
"/*/*[@class='header_completed'][1]
/following-sibling::*
[count(.|/*/*[@class='header_completed'][1]
/following-sibling::*[@align='center'][1]
/preceding-sibling::*)
=
count(/*/*[@class='header_completed'][1]
/following-sibling::*[@align='center'][1]
/preceding-sibling::*)
]
"/>
</xsl:template>
</xsl:stylesheet>
<table border="0" cellpadding="0" cellspacing="1" width="920"/>
<table border="0" cellpadding="0" cellspacing="2" width="920"/>
<table border="0" cellpadding="0" cellspacing="3" width="920"/>
<table border="0" cellpadding="0" cellspacing="4" width="920"/>
<table border="0" cellpadding="0" cellspacing="5" width="920"/>
<DELIMITER/>
<table border="0" cellpadding="0" cellspacing="1" width="920"/>
<table border="0" cellpadding="0" cellspacing="2" width="920"/>
<table border="0" cellpadding="0" cellspacing="3" width="920"/>
<table border="0" cellpadding="0" cellspacing="4" width="920"/>
<table border="0" cellpadding="0" cellspacing="5" width="920"/>
两个节点集$ns1
和$ns2
的交集由以下XPath表达式计算：
$ns1[count(.| $ns2)=count($ns2)]
<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:variable name="ns1" select=
"/*/*[ .
>>
/*/*[@class='header_completed'][1]
]
"/>
<xsl:variable name="ns2" select=
"/*/*[ /*/*[@class='header_completed'][1]
/following-sibling::*[@align='center'][1]
>>
.
]
"/>
<xsl:template match="/">
<xsl:sequence select="$ns1 intersect $ns2"/>
<DELIMITER/>
<xsl:sequence select=
"/*/*[ .
>>
/*/*[@class='header_completed'][1]
]
intersect
/*/*[ /*/*[@class='header_completed'][1]
/following-sibling::*[@align='center'][1]
>>
.
]
"/>
</xsl:template>
</xsl:stylesheet>
如果我们有以下XML文档：
<t>
<table border="0" cellpadding="0" cellspacing="0" width="920" align="center"></table>
<table border="0" cellpadding="0" cellspacing="0" width="920"></table>
<table border="0" cellpadding="0" cellspacing="0" width="920"></table>
<table border="0" cellpadding="0" cellspacing="0" width="920" align="center" class="header_completed"></table>
<table border="0" cellpadding="0" cellspacing="1" width="920"></table>
<table border="0" cellpadding="0" cellspacing="2" width="920"></table>
<table border="0" cellpadding="0" cellspacing="3" width="920"></table>
<table border="0" cellpadding="0" cellspacing="4" width="920"></table>
<table border="0" cellpadding="0" cellspacing="5" width="920"></table>
<table border="0" cellpadding="0" cellspacing="0" width="920" align="center"></table>
<table border="0" cellpadding="0" cellspacing="0" width="920"></table>
<table border="0" cellpadding="0" cellspacing="0" width="920"></table>
<table border="0" cellpadding="0" cellspacing="0" width="920" align="center"></table>
</t>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:variable name="ns1" select=
"/*/*[@class='header_completed'][1]
/following-sibling::*
"/>
<xsl:variable name="ns2" select=
"/*/*[@class='header_completed'][1]
/following-sibling::*[@align='center'][1]
/preceding-sibling::*
"/>
<xsl:template match="/">
<xsl:copy-of select=
"$ns1[count(.| $ns2)=count($ns2)]
"/>
<DELIMITER/>
<xsl:copy-of select=
"/*/*[@class='header_completed'][1]
/following-sibling::*
[count(.|/*/*[@class='header_completed'][1]
/following-sibling::*[@align='center'][1]
/preceding-sibling::*)
=
count(/*/*[@class='header_completed'][1]
/following-sibling::*[@align='center'][1]
/preceding-sibling::*)
]
"/>
</xsl:template>
</xsl:stylesheet>
<table border="0" cellpadding="0" cellspacing="1" width="920"/>
<table border="0" cellpadding="0" cellspacing="2" width="920"/>
<table border="0" cellpadding="0" cellspacing="3" width="920"/>
<table border="0" cellpadding="0" cellspacing="4" width="920"/>
<table border="0" cellpadding="0" cellspacing="5" width="920"/>
<DELIMITER/>
<table border="0" cellpadding="0" cellspacing="1" width="920"/>
<table border="0" cellpadding="0" cellspacing="2" width="920"/>
<table border="0" cellpadding="0" cellspacing="3" width="920"/>
<table border="0" cellpadding="0" cellspacing="4" width="920"/>
<table border="0" cellpadding="0" cellspacing="5" width="920"/>
$ns2
是：
/*/*[@class='header_completed'][1]
/following-sibling::*
/*/*[@class='header_completed'][1]
/following-sibling::*[@align='center'][1]
/preceding-sibling::*
我们只需在Kayessian公式中替换$ns1
和$ns2
，即可得到以下XPath表达式，该表达式正好选择所需的5个元素：
/*/*[@class='header_completed'][1]
/following-sibling::*
[count(.|/*/*[@class='header_completed'][1]
/following-sibling::*[@align='center'][1]
/preceding-sibling::*)
=
count(/*/*[@class='header_completed'][1]
/following-sibling::*[@align='center'][1]
/preceding-sibling::*)
]
为了验证这是真正的解决方案，我们使用以下XSLT转换：
<t>
<table border="0" cellpadding="0" cellspacing="0" width="920" align="center"></table>
<table border="0" cellpadding="0" cellspacing="0" width="920"></table>
<table border="0" cellpadding="0" cellspacing="0" width="920"></table>
<table border="0" cellpadding="0" cellspacing="0" width="920" align="center" class="header_completed"></table>
<table border="0" cellpadding="0" cellspacing="1" width="920"></table>
<table border="0" cellpadding="0" cellspacing="2" width="920"></table>
<table border="0" cellpadding="0" cellspacing="3" width="920"></table>
<table border="0" cellpadding="0" cellspacing="4" width="920"></table>
<table border="0" cellpadding="0" cellspacing="5" width="920"></table>
<table border="0" cellpadding="0" cellspacing="0" width="920" align="center"></table>
<table border="0" cellpadding="0" cellspacing="0" width="920"></table>
<table border="0" cellpadding="0" cellspacing="0" width="920"></table>
<table border="0" cellpadding="0" cellspacing="0" width="920" align="center"></table>
</t>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:variable name="ns1" select=
"/*/*[@class='header_completed'][1]
/following-sibling::*
"/>
<xsl:variable name="ns2" select=
"/*/*[@class='header_completed'][1]
/following-sibling::*[@align='center'][1]
/preceding-sibling::*
"/>
<xsl:template match="/">
<xsl:copy-of select=
"$ns1[count(.| $ns2)=count($ns2)]
"/>
<DELIMITER/>
<xsl:copy-of select=
"/*/*[@class='header_completed'][1]
/following-sibling::*
[count(.|/*/*[@class='header_completed'][1]
/following-sibling::*[@align='center'][1]
/preceding-sibling::*)
=
count(/*/*[@class='header_completed'][1]
/following-sibling::*[@align='center'][1]
/preceding-sibling::*)
]
"/>
</xsl:template>
</xsl:stylesheet>
<table border="0" cellpadding="0" cellspacing="1" width="920"/>
<table border="0" cellpadding="0" cellspacing="2" width="920"/>
<table border="0" cellpadding="0" cellspacing="3" width="920"/>
<table border="0" cellpadding="0" cellspacing="4" width="920"/>
<table border="0" cellpadding="0" cellspacing="5" width="920"/>
<DELIMITER/>
<table border="0" cellpadding="0" cellspacing="1" width="920"/>
<table border="0" cellpadding="0" cellspacing="2" width="920"/>
<table border="0" cellpadding="0" cellspacing="3" width="920"/>
<table border="0" cellpadding="0" cellspacing="4" width="920"/>
<table border="0" cellpadding="0" cellspacing="5" width="920"/>
下面是一个XSLT 2.0解决方案，证明了此XSLT 2.0表达式的正确性：
$ns1[count(.| $ns2)=count($ns2)]
<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:variable name="ns1" select=
"/*/*[ .
>>
/*/*[@class='header_completed'][1]
]
"/>
<xsl:variable name="ns2" select=
"/*/*[ /*/*[@class='header_completed'][1]
/following-sibling::*[@align='center'][1]
>>
.
]
"/>
<xsl:template match="/">
<xsl:sequence select="$ns1 intersect $ns2"/>
<DELIMITER/>
<xsl:sequence select=
"/*/*[ .
>>
/*/*[@class='header_completed'][1]
]
intersect
/*/*[ /*/*[@class='header_completed'][1]
/following-sibling::*[@align='center'][1]
>>
.
]
"/>
</xsl:template>
</xsl:stylesheet>
当应用于之前定义的XML文档时，我们再次得到了相同的正确结果：
<table border="0" cellpadding="0" cellspacing="1" width="920"/>
<table border="0" cellpadding="0" cellspacing="2" width="920"/>
<table border="0" cellpadding="0" cellspacing="3" width="920"/>
<table border="0" cellpadding="0" cellspacing="4" width="920"/>
<table border="0" cellpadding="0" cellspacing="5" width="920"/>
<DELIMITER/>
<table border="0" cellpadding="0" cellspacing="1" width="920"/>
<table border="0" cellpadding="0" cellspacing="2" width="920"/>
<table border="0" cellpadding="0" cellspacing="3" width="920"/>
<table border="0" cellpadding="0" cellspacing="4" width="920"/>
<table border="0" cellpadding="0" cellspacing="5" width="920"/>
我相信这个XPath表达式会选择您想要的节点：
//table[@class="header_completed"]/
following-sibling::table[@align="center"][1]/
preceding-sibling::table[
preceding-sibling::table[@class="header_completed"]
]
首先，我使用@class=“header\u completed”
导航到表
从那里，我用@align=“center”
选择下面的第一个同级表
从那里，我选择所有前面的同级表，这些表都有一个前面的同级表，即带有@class=“header\u completed”
好问题（+1）的表。请看我的答案以获得解释和完整的解决方案。该死，你真的尽了最大努力去问一个好的棘手问题。关于实用性和鲁棒性设计的完整要点，我非常想将其扩展到[@class='header\u completed']
…一些节点的两个实例…[@align='center']
，并一次性捕获这两个集合。+1用于实际可行的解决方案，无需硬编码位置，思维良好。然而，在扩展它之后，我想知道你是否会对我在对问题的评论中提出的可能性有任何见解：如果我们在节点之间有两个有效的序列会怎么样，即：如果我们在自身之后复制示例，如何仍然得到10个结果节点（在match1\u of u start
和match1\u of u end
+match2\u of u start
和match2\u of u end
）之间，而不是整个列表（在match1\u of u start
和match2\u of u end
）.+1这是一个很好的解决方案，也很容易理解。非常感谢。@Wrikken这肯定是一个挑战。我必须考虑这个问题。+1这是一个非常有洞察力的答案，也是一个非常有用的答案。它有很好的解释和证明。谢谢。+1这是这个问题的实际答案，也是凯西公式A的一个有指导意义的例子应用程序也是如此。