How to check if <a href> element exist in a <div> element?

user_5541517 0
Viewed 1665

the html code is like this:

<div class="AAA">Text of AAA<a href="......AAA/url">Display text of URL A</a></div>
<div class="BBB">Text of BBB<a href="......BBB/url">Display text of URL B</a></div>
<div class="CCC">Text of CCC</div>
<div class="DDD">Text of DDD</div>

I want to parse the text for all the div, while check if there is url exist, if yes then also extract it out and display in output

output like this:

Text of AAA
Display text of URL A
......AAA/url
Text of BBB
Display text of URL B
......BBB/url
Text of CCC
Text of DDD

i tried to nest the loop of find_all('a') within find_all('div') loop, but messed up my output

beautifulsoup python
5 Answers
from bs4 import BeautifulSoup
html="""
<div class="AAA">Text of AAA<a href="......AAA/url">Display text of URL A</a></div>
<div class="BBB">Text of BBB<a href="......BBB/url">Display text of URL B</a></div>
<div class="CCC">Text of CCC</div>
<div class="DDD">Text of DDD</div>
"""
soup = BeautifulSoup(html, "lxml")
for div in soup.findAll('div'):
    print(div.text)
    try:
        print(div.find('a').text)
        print(div.find('a')["href"])
    except AttributeError:
        pass

Output

Text of AAADisplay text of URL A
Display text of URL A
......AAA/url
Text of BBBDisplay text of URL B
Display text of URL B
......BBB/url
Text of CCC
Text of DDD
user_108255560

Don't know what your code looks like, but the basic idea is something like this:

data = soup.findAll('div')
for div in data:
    links = div.findAll('a')
    for a in links:
        print(a['href'])
        print(a.text)

will give you the URL and the text.

user_62207590

You can loop over the divs and then print the elements of soup.contents:

s = """
<div class="AAA">Text of AAA<a href="......AAA/url">Display text of URL A</a> . 
</div>
<div class="BBB">Text of BBB<a href="......BBB/url">Display text of URL B</a> . 
</div>
<div class="CCC">Text of CCC</div>
<div class="DDD">Text of DDD</div>
"""
from bs4 import BeautifulSoup as soup
for _text, *_next in map(lambda x:x.contents, soup(s, 'html.parser').find_all('div')):
  print(_text)
  if _next:
    print(_next[0].text)
    print(_next[0]['href'])

Output:

Text of AAA
Display text of URL A
......AAA/url
Text of BBB
Display text of URL B
......BBB/url
Text of CCC
Text of DDD
user_73267380

It just easier to read, You can also use this to get expected output

divs = soup.find_all('div')
for div in divs:
  print(div.contents[0]) # Text of AAA
  link = div.find('a')
  if link:
    print(link.text) # Display text of URL A
    print(link['href']) # ......AAA/url
user_4582140

thanks all, i worked out the solution

for h in ans_kin:
    links = ""
    link = h.find('a')
    if link:
        for l in link:
            links = h.text + link.get('href')
    else:
        links = h.text

    answer_kin.append(links)
user_55415170
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