Question

我有一个html文件，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，如果所有标签都在＆＃39; p＆＃39;块是强大的，然后将它们组合成一行，即

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

不触及另一个块，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，我试过了：

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，我能够剥离所需部分中的强标记，并给出：

<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

我能用bs4（BeautifulSoup）做到这一点：

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

我设法解决了自己的问题。

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，他帮我解决了这个问题。虽然我不能正确地回答他的答案，但我对他的指导并不那么感激。

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

关于正在使用的xpath的简要说明：

//p[strong]：在XML / HTML文档中的任何位置找到p元素，具有子元素strong ...
[not(*[not(self::strong)])]：..并且没有strong ...
[not(text()[normalize-space()])]：..并且没有非空的文本节点子。
normalize-space()：从当前上下文元素获取所有文本节点，与标准化为单个空格的连续空格连接

将多个标签与lxml组合使用

3 个答案: