将多个标签与 lxml 组合

Question

我有一个 html 文件，它看起来像：

...
<p>  
    <strong>This is </strong>  
    <strong>a lin</strong>  
    <strong>e which I want to </strong>  
    <strong>join.</strong>  
</p>
<p>
    2.
    <strong>But do not </strong>
    <strong>touch this</strong>
    <em>Maybe some other tags as well.</em>
    bla bla blah...
</p>
...

我需要的是，如果一个'p'块中的所有标签都是'strong'，那么将它们合并成一行，即

<p>
    <strong>This is a line which I want to join.</strong>
</p>

不接触另一个方块，因为它包含其他内容。

有什么建议吗？我正在使用 lxml。

更新：

到目前为止我试过：

for p in self.tree.xpath('//body/p'):
        if p.tail is None: #no text before first element
            children = p.getchildren()
            for child in children:
                if len(children)==1 or child.tag!='strong' or child.tail is not None:
                    break
            else:
                etree.strip_tags(p,'strong')

使用这些代码，我能够去除所需部分中的强标签，给出：

<p>
      This is a line which I want to join.  
</p>

所以现在我只需要一种方法将标签放回...

Answer 1

我可以用 bs4 做到这一点 (BeautifulSoup):

from bs4 import BeautifulSoup as bs

html = """<p>  
<strong>This is </strong>  
<strong>a lin</strong>  
<strong>e which I want to </strong>  
<strong>join.</strong>  
</p>
<p>
<strong>But do not </strong>
<strong>touch this</strong>
</p>"""

soup = bs(html)
s = ''
# note that I use the 0th <p> block ...[0],
# so make the appropriate change in your code
for t in soup.find_all('p')[0].text:
    s = s+t.strip('\n')
s = '<p><strong>'+s+'</strong></p>'
print s # prints: <p><strong>This is a line which I want to join.</strong></p>

然后使用replace_with():

p_tag = soup.p
p_tag.replace_with(bs(s, 'html.parser'))
print soup

打印：

<html><body><p><strong>This is a line which I want to join.</strong></p>
<p>
<strong>But do not </strong>
<strong>touch this</strong>
</p></body></html>

Answer 2

我已经设法解决了我自己的问题。

for p in self.tree.xpath('//body/p'):
    if p.tail is None:  # some conditions specifically for my doc 
        children = p.getchildren()
        if len(children)>1:
            for child in children:
                #if other stuffs present, break
                if child.tag!='strong' or child.tail is not None: 
                    break
            else:
                # If not break, we find a p block to fix
                # Get rid of stuffs inside p, and put a SubElement in
                etree.strip_tags(p,'strong')
                tmp_text = p.text_content()
                p.clear()
                subtext = etree.SubElement(p, "strong")
                subtext.text = tmp_text

特别感谢@Scott 帮助我得出这个解决方案。虽然我不能标记他的答案是正确的，但我对他的指导表示赞赏。

Answer 3

或者，您可以使用更具体的 xpath 直接获取目标 p 元素：

p_target = """
//p[strong]
   [not(*[not(self::strong)])]
   [not(text()[normalize-space()])]
"""
for p in self.tree.xpath(p_target):
    #logic inside the loop can also be the same as your `else` block
    content = p.xpath("normalize-space()")
    p.clear()
    strong = etree.SubElement(p, "strong")
    strong.text = content

使用xpath的简要说明：

//p[strong] : 查找 p 元素，在 XML/HTML 文档中的任何位置，具有子元素 strong...
[not(*[not(self::strong)])] : ..除了 strong...
[not(text()[normalize-space()])] : ..并且没有非空文本节点子节点。
normalize-space() ：从当前上下文元素中获取所有文本节点，并与标准化为单个 space

将多个标签与 lxml 组合

Combine multiple tags with lxml

html

python

xpath

lxml