让Nokogiri选择start和stop元素之间的所有内容(包括start / stop-element)最聪明的方法是什么?
检查下面的示例代码,了解我在寻找的内容:
require 'rubygems'
require 'nokogiri'
value = Nokogiri::HTML.parse(<<-HTML_END)
"<html>
<body>
<p id='para-1'>A</p>
<div class='block' id='X1'>
<p class="this">Foo</p>
<p id='para-2'>B</p>
</div>
<p id='para-3'>C</p>
<p class="that">Bar</p>
<p id='para-4'>D</p>
<p id='para-5'>E</p>
<div class='block' id='X2'>
<p id='para-6'>F</p>
</div>
<p id='para-7'>F</p>
<p id='para-8'>G</p>
</body>
</html>"
HTML_END
parent = value.css('body').first
# START element
@start_element = parent.at('p#para-3')
# STOP element
@end_element = parent.at('p#para-7')
结果(返回值)应该如下所示:
<p id='para-3'>C</p>
<p class="that">Bar</p>
<p id='para-4'>D</p>
<p id='para-5'>E</p>
<div class='block' id='X2'>
<p id='para-6'>F</p>
</div>
<p id='para-7'>F</p>
更新:这是我目前的解决方案,不过我觉得必须有更聪明的东西:
@my_content = ""
@selected_node = true
def collect_content(_start)
if _start == @end_element
@my_content << _start.to_html
@selected_node = false
end
if @selected_node == true
@my_content << _start.to_html
collect_content(_start.next)
end
end
collect_content(@start_element)
puts @my_content
答案 0 :(得分:10)
使用递归的一种非常聪明的oneliner:
def collect_between(first, last)
first == last ? [first] : [first, *collect_between(first.next, last)]
end
迭代解决方案:
def collect_between(first, last)
result = [first]
until first == last
first = first.next
result << first
end
result
end
编辑:(简短)对星号的解释
它被称为splat运算符。它“展开”一个数组:
array = [3, 2, 1]
[4, array] # => [4, [3, 2, 1]]
[4, *array] # => [4, 3, 2, 1]
some_method(array) # => some_method([3, 2, 1])
some_method(*array) # => some_method(3, 2, 1)
def other_method(*array); array; end
other_method(1, 2, 3) # => [1, 2, 3]
答案 1 :(得分:2)
# monkeypatches for Nokogiri::NodeSet
# note: versions of these functions will be in Nokogiri 1.3
class Nokogiri::XML::NodeSet
unless method_defined?(:index)
def index(node)
each_with_index { |member, j| return j if member == node }
end
end
unless method_defined?(:slice)
def slice(start, length)
new_set = Nokogiri::XML::NodeSet.new(self.document)
length.times { |offset| new_set << self[start + offset] }
new_set
end
end
end
#
# solution #1: picking elements out of node children
# NOTE that this will also include whitespacy text nodes between the <p> elements.
#
possible_matches = parent.children
start_index = possible_matches.index(@start_element)
stop_index = possible_matches.index(@end_element)
answer_1 = possible_matches.slice(start_index, stop_index - start_index + 1)
#
# solution #2: picking elements out of a NodeSet
# this will only include elements, not text nodes.
#
possible_matches = value.xpath("//body/*")
start_index = possible_matches.index(@start_element)
stop_index = possible_matches.index(@end_element)
answer_2 = possible_matches.slice(start_index, stop_index - start_index + 1)
答案 2 :(得分:2)
为了完整起见,仅限XPath 解决方案:)
它构建了两个集合的交集,即start元素的以下兄弟节点和结束元素的前面兄弟节点。
基本上你可以用:
建立交集$a[count(.|$b) = count($b)]
对可读性的变量略有不同:
@start_element = "//p[@id='para-3']"
@end_element = "//p[@id='para-7']"
@set_a = "#@start_element/following-sibling::*"
@set_b = "#@end_element/preceding-sibling::*"
@my_content = value.xpath("#@set_a[ count(.|#@set_b) = count(#@set_b) ]
| #@start_element | #@end_element")
兄弟姐妹不包含元素本身,因此开始和结束元素必须单独包含在表达式中。
修改:更简单的解决方案:
@start_element = "p[@id='para-3']"
@end_element = "p[@id='para-7']"
@my_content = value.xpath("//*[preceding-sibling::#@start_element and
following-sibling::#@end_element]
| //#@start_element | //#@end_element")