How can you parse reorder links using Beautiful Soup?

S

Sergey2020-10-21 10:45:49

beautiful soup

Sergey, 2020-10-21 10:45:49

===================================================
<html>
<div class="item">
  <div class="quality">SD</div>
  <div class="link"><a href>"http://link"</div>
<div class="item">
  <div class="quality">HD</div>
  <div class="link"><a href>"http://link"</div>
<div class="item">
  <div class="quality">FullHD</div>
  <div class="link"><a href>"http://link"</div>
</html>
===================================================
<html>
<div class="item">
  <div class="quality">SD</div>
  <div class="link"><a href>"http://link"</div>
<div class="item">
  <div class="quality">FullHD</div>
  <div class="link"><a href>"http://link"</div>
<div class="item">
  <div class="quality">HD</div>
  <div class="link"><a href>"http://link"</div>
</html>
===================================================
<html>
<div class="item">
  <div class="quality">FullHD</div>
  <div class="link"><a href>"http://link"</div>
<div class="item">
  <div class="quality">SD</div>
  <div class="link"><a href>"http://link"</div>
<div class="item">
  <div class="quality">HD</div>
  <div class="link"><a href>"http://link"</div>
</html>
===================================================

How to get the correct links from the desired "quality" class if the output sequence changes with each request above gave an example of the output that needs to be parsed

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2 answer(s)

S

soremix, 2020-10-21
@soremix

And how do you understand that the right link is in front of you? Look at the name of the quality, and the link below it? Why should other logic be in the code?

sources = {}
for item in items:
    quality = item.find('div', {'class': 'quality'}).text
    link = item.find('div', {'class': 'link'}).text
    sources[quality] = link

D

datka, 2020-10-21
@datka

Loop parse and find links Or parse directly BS4 Documentation<div class="item"><a href>"http://link"