Extract src in beatifulsoup parser python

Question

need extract src in tag with beatifulsoup in python of tag 'div.participant-logo' .

import requests
import bs4

root_url = 'here url to parse'

response = requests.get(root_url)
soup = bs4.BeautifulSoup(response.text)

logo_cuadro1 = soup.select('div.participant-logo')

print (logo_cuadro1)

Code HTML

<td class="participant-logo"><a href="/futbol/espana/equipo-cordoba-cf-8004992.html"><img src="http://medias/logos/icons/teams-80/7869.png?v=2"></a></td>

Don't use `response.text`. Use `response.content`, perhaps with `response.encoding` passed in to `BeautifulSoup` as the `from_encoding` argument, provided `'charset' in response.headers.get('content-type')` is true. See [retrieve links from web page using python and BeautifulSoup](http://stackoverflow.com/a/22583436) — Martijn Pieters, Jan 16 '15 at 19:43
`logo_cuadro1 = soup.select('div.participant-logo').find("src")` — Padraic Cunningham, Jan 16 '15 at 19:43
What is the actual HTML you have? What is the expected output? What do you mean by *src in tag*? — Martijn Pieters, Jan 16 '15 at 19:43
@PadraicCunningham: that's not clear; perhaps the OP wanted the source of the whole tag as HTML? — Martijn Pieters, Jan 16 '15 at 19:44
@MartijnPieters, well the question is pretty unclear, a literal interpretation of the question suggests what I provided. — Padraic Cunningham, Jan 16 '15 at 19:46
@PadraicCunningham: which is why I voted to close it. As it stands it is unanswerable. — Martijn Pieters, Jan 16 '15 at 19:49
@Fabián, added an answer, I presume you want the link from src? — Padraic Cunningham, Jan 16 '15 at 20:00

Padraic Cunningham · Accepted Answer · 2015-01-16T20:09:34.877

0

To get the src:

soup = bs4.BeautifulSoup("""<td class="participant-logo"><a href="/futbol/espana/equipo-cordoba-cf-8004992.html"><img src="http://medias/logos/icons/teams-80/7869.png?v=2"></a></td>""")

logo_cuadro1 = soup.find("img")["src"]

print (logo_cuadro1)

http://medias/logos/icons/teams-80/7869.png?v=2

In your case:

sel  = soup.select('div.participant-logo')
link = sel[0].find("img")["src"]

edited Jan 16 '15 at 20:09

answered Jan 16 '15 at 19:58

Padraic Cunningham

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Hi, this return error. logo_cuadro1 = soup.select('div.participant-logo').find("img")["src"] AttributeError: 'list' object has no attribute 'find' – Fabián Jan 16 '15 at 20:02

score 0 · Answer 2 · answered Jan 16 '15 at 20:07

0

You can extend your CSS selection to the an image with src attribute:

for img in soup.select('div.participant-logo img[src]'):
    print img['src']

answered Jan 16 '15 at 20:07

Martijn Pieters

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Extract src in beatifulsoup parser python

2 Answers2