How does the Python crawler crawl the content between span and span and store them in the dictionary respectively?

Question

I want to capture the house profiles separately and store them in the dictionary as independent columns, but there is no way to directly extract the inline elements using a for loop.
This is my code:

soup.select('.house-info li')[1].text.strip()

This is the html code of the web page:

<li><span class="info-tit">房屋概况：</span>住宅
                        <span class="splitline">|</span>1室1厅1卫
                        
                        <span class="splitline">|</span><span title="建筑面积">46m²</span>
                        
                        <span class="splitline">|</span> (高层)/共18层
                        
                        <span class="splitline">|</span>南北
                        
                        <span class="splitline">|</span> 豪华装修
                        
                    </li>

Answer 1

Actually, it is very simple. You can see that there is a pattern in this. The pattern lies in the separator |. I wrote a DEMO

something  = '''<li><span class="info-tit">房屋概况：</span>住宅  <span class="splitline">|</span>1室1厅1卫<span class="splitline">|</span><span title="建筑面积">46m²</span><span class="splitline">|</span> (高层)/共18层

                        <span class="splitline">|</span>南北

                        <span class="splitline">|</span> 豪华装修

                    </li>''';

soup  = BeautifulSoup(something, 'lxml')
plaintext = soup.select('li')[0].get_text().strip()

Get all the inner content through get_text(), and then remove the spaces. You can use split to divide it later, and I won’t write the rest.
If you have any questions, please communicate.

Answer 2

I feel that this html code is written wrong, the content text of the label is outside the label

There are only two correct label contents:

• House Overview:

• 46m²

Answer 3

innerText

Answer 4

In your case, I think it is most convenient to use a for loop plus regular expressions, if all templates are fixed like this

Answer 5

用pyquery吧

from pyquery import PyQuery as Q

Q(text).find('.house-info li').text()