不同类下同名，获取URL，BeautifulSoup Python答案

【问题标题】：Same name under different class, get URL, BeautifulSoup Python不同类下同名，获取URL，BeautifulSoup Python
【发布时间】：2020-10-08 21:57:31
【问题描述】：

我有这个 HTML，我需要获取上面的 URL：

<div class="posts-container col-md-6"
   <ul class="emb-embassies-list"
     <a class="entry-title" href="commonlink.com"
   <ul class="emb-embassies-list"
     <a class="entry-title" href="rarelink.com"

<div class="col-md-6"
   <ul class="emb-embassies-list"
     <a class="entry-title" href="anothercommonlink.com"
   <ul class="emb-embassies-list"
     <a class="entry-title" href="legendarylink.com"

当我申请时：

for i in soup.findAll('div', "posts-container col-md-6"):
    for anchor in soup.findAll('a', class_="entry-title", href=True):
        print(anchor['href'])

我明白了：

>commonlink.com
>rarelink.com
>anothercommonlink.com
>legendarylink.com

我只想获取“posts-container col-md-6”：

>commonlink.com
>rarelink.com

【问题讨论】：

在内部循环中迭代i，而不是整个页面for anchor in i.findAll('a', class_="entry-title", href=True)。
@Sushanth 出于某种原因，它只返回“进程以退出代码 0 完成”：/

标签： python html url beautifulsoup

【解决方案1】：

要使用class="posts-container col-md-6" 获取<div> 下的所有链接，请使用CSS 选择器.posts-container.col-md-6 a：

from bs4 import BeautifulSoup


txt = '''
<div class="posts-container col-md-6">
   <ul class="emb-embassies-list">
     <a class="entry-title" href="commonlink.com">some link1</a>
   <ul class="emb-embassies-list">
     <a class="entry-title" href="rarelink.com">some link2</a>
</div>
<div class="col-md-6">
   <ul class="emb-embassies-list">
     <a class="entry-title" href="anothercommonlink.com">some link3</a>
   <ul class="emb-embassies-list">
     <a class="entry-title" href="legendarylink.com">some link4</a>
</div>'''



soup = BeautifulSoup(txt, 'html.parser')

for a in soup.select('.posts-container.col-md-6 a'):
    print(a['href'])

打印：

commonlink.com
rarelink.com

【讨论】：

【解决方案2】：

你可以试试：

from bs4 import BeautifulSoup

html_doc = '''
<div class="posts-container col-md-6">
   <ul class="emb-embassies-list">
     <a class="entry-title" href="commonlink.com">some link1</a>
   <ul class="emb-embassies-list">
     <a class="entry-title" href="rarelink.com">some link2</a>
</div>
<div class="col-md-6">
   <ul class="emb-embassies-list">
     <a class="entry-title" href="anothercommonlink.com">some link3</a>
   <ul class="emb-embassies-list">
     <a class="entry-title" href="legendarylink.com">some link4</a>
</div>'''

soup = BeautifulSoup(html_doc, 'lxml')
ancors = soup.find('div', class_="posts-container col-md-6").find_all('a')

for a in ancors:
    print(a['href'])

输出将是：

commonlink.com
rarelink.com

【讨论】：