Develop Reference

Unable to find element BeautifulSoup

I am trying to parse a specific href link from the following website: https://www.murray-intl.co.uk/en/literature-library.
Element i seek to parse:
<a class="btn btn--naked btn--icon-left btn--block focus-within" href="https://www.aberdeenstandard.com/docs?editionId=9123afa2-5318-4715-9783-e07d08e2e7cc&_ga=2.12911351.1364356977.1629796255-1577053129.1629192717" target="blank">Portfolio Holding Summary<i class="material-icons btn__icon">library_books</i></a>
However, using BeautifulSoup I am unable to obtain the desired element, perhaps due to cookies acceptance.
from bs4 import BeautifulSoup
import urllib.request
import requests as rq
page = requests.get('https://www.murray-intl.co.uk/en/literature-library')
soup = BeautifulSoup(page.content, 'html.parser')
link = soup.find_all('a', class_='btn btn--naked btn--icon-left btn--block focus-within')
url = link[0].get('href')
url
I am still new at BS4, and hope someone can help me on the right course.
Thank you in advance!

To get correct tags, remove "focus-within" class (it's added later by JavaScript):
import requests
from bs4 import BeautifulSoup
url = "https://www.murray-intl.co.uk/en/literature-library"
soup = BeautifulSoup(requests.get(url).content, "html.parser")
links = soup.find_all("a", class_="btn btn--naked btn--icon-left btn--block")
for u in links:
print(u.get_text(strip=True), u.get("href", ""))
Prints:
...
Portfolio Holding Summarylibrary_books https://www.aberdeenstandard.com/docs?editionId=9123afa2-5318-4715-9783-e07d08e2e7cc
...
EDIT: To get only the specified link you can use for example CSS selector:
link = soup.select_one('a:-soup-contains("Portfolio Holding Summary")')
print(link["href"])
Prints:
https://www.aberdeenstandard.com/docs?editionId=9123afa2-5318-4715-9783-e07d08e2e7cc

Can't extract src attribute from "img" tag with BeautifulSoup

I'm working on a project and I'm trying to extract the pictures' URL from a website. I'm a noob at this so please bear with me. Based on the HTML code, the class of the pictures that I want is "fotorama__img". However, when I execute my code, it doesn't seem to work. Anyone knows why that's the case? Also, how come the src attribute doesn't contain the whole URL, just a part of it? Example: the link to the image is https://www.supermicro.com/files_SYS/images/System/SYS-120U-TNR_callout_front.jpg but the src attribute of the img tag is "/files_SYS/images/System/sysThumb/SYS-120U-TNR_main.png".
Here is my code:
from bs4 import BeautifulSoup
import requests
page = requests.get("https://www.supermicro.com/en/products/system/Ultra/1U/SYS-120U-TNR")
soup = BeautifulSoup(page.content,'lxml')
images = soup.find_all("img", {"class": "fotorama__img"})
for image in images:
print(image.get("src"))
And here is the picture of the HTML code for the page
Thank you for your help!

The class is added dynamically via JavaScript, so beautifulsoup doesn't see it. To extract the images from this site, you can do:
import requests
from bs4 import BeautifulSoup
page = requests.get(
"https://www.supermicro.com/en/products/system/Ultra/1U/SYS-120U-TNR"
)
soup = BeautifulSoup(page.content, "lxml")
images = [
"https://www.supermicro.com" + a["href"]
for a in soup.select(".fotorama > a")
]
print(*images, sep="\n")
Prints:
https://www.supermicro.com/files_SYS/images/System/SYS-120U-TNR_main.png
https://www.supermicro.com/files_SYS/images/System/SYS-120U-TNR_callout_angle.jpg
https://www.supermicro.com/files_SYS/images/System/SYS-120U-TNR_callout_top.jpg
https://www.supermicro.com/files_SYS/images/System/SYS-120U-TNR_callout_front.jpg
https://www.supermicro.com/files_SYS/images/System/SYS-120U-TNR_callout_rear.jpg

Python Beautiful soup: Target a particular element

I am trying to scrape a particular part of a website(https://flightmath.com/from-CDG-to-BLR) but I am unable to target the element that I need.
Below is the part of the html
<h2 style="background-color:#7DC2F8;padding:10px"><i class="fa fa-plane"></i>
flight distance = <strong>4,866</strong> miles</h2>
This is my code
dist = soup.find('h2', attrs={'class': 'fa fa-plane'})
I just want to target the "4,866" part.
I would be really grateful if someone can guide me on this.
Thanks in advance.

attrs={'class': '...'} requires an exact class attribute value (not a combination). Instead, use soup.select_one method to select by extended css rule:
from bs4 import BeautifulSoup
import requests
url = 'https://flightmath.com/from-CDG-to-BLR'
html_data = requests.get(url).content
soup = BeautifulSoup(html_data, 'html.parser')
dist = soup.select_one('h2 i.fa-plane + strong')
print(dist.text) # 4,866

In case of interest: The value is hard coded into the html (for a flight speed calculation) so you could also regex out a more precise value with the following. You can use round() to get the value shown on page.
import requests, re
urls = ['https://flightmath.com/from-CDG-to-BOM', 'https://flightmath.com/from-CDG-to-BLR', 'https://flightmath.com/from-CDG-to-IXC']
p = re.compile(r'flightspeed\.min\.value\/60 \+ ([0-9.]+)')
with requests.Session() as s:
for url in urls:
print(p.findall(s.get(url).text)[0])

find tag with class name and then use find_next() to find the strong tag.
from bs4 import BeautifulSoup
import requests
url = 'https://flightmath.com/from-CDG-to-BLR'
html_data = requests.get(url).text
soup = BeautifulSoup(html_data, 'html.parser')
dist = soup.find('i',class_='fa-plane').find_next('strong')
print(dist.text)

How to scrape src from img html in python

I'm trying to scrape the src of the img, but the code I found returns many img src, but not the one I want. I can't figure out what I am doing wrong. I am scraping TripAdvisor on "https://www.tripadvisor.dk/Restaurant_Review-g189541-d15804886-Reviews-The_Pescatarian-Copenhagen_Zealand.html"
So this is the HTML snippet I'm trying to extract from:
<div class="restaurants-detail-overview-cards-LocationOverviewCard__cardColumn--2ALwF"><h6>Placering og kontaktoplysninger</h6><span><div><span data-test-target="staticMapSnapshot" class=""><img class="restaurants-detail-overview-cards-LocationOverviewCard__mapImage--22-Al" src="https://trip-raster.citymaps.io/staticmap?scale=1&zoom=15&size=347x137&language=da&center=55.687988,12.596316&markers=icon:http%3A%2F%2Fc1.tacdn.com%2F%2Fimg2%2Fmaps%2Ficons%2Fcomponent_map_pins_v1%2FR_Pin_Small.png|55.68799,12.596316"></span></div></span>
I want the code to return: (a sub-string from src)
55.68799,12.596316
I have tried:
import pandas as pd
pd.options.display.max_colwidth = 200
from urllib.request import urlopen
from bs4 import BeautifulSoup as bs
import re
web_url = "https://www.tripadvisor.dk/Restaurant_Review-g189541-d15804886-Reviews-The_Pescatarian-Copenhagen_Zealand.html"
url = urlopen(web_url)
url_html = url.read()
soup = bs(url_html, 'lxml')
soup.find_all('img')
for link in soup.find_all('img'):
print(link.get('src'))
the return is along the lines of this BUT NOT the src that I need :
https://static.tacdn.com/img2/branding/rebrand/TA_logo_secondary.svg
https://static.tacdn.com/img2/branding/rebrand/TA_logo_primary.svg
https://static.tacdn.com/img2/branding/rebrand/TA_logo_secondary.svg
data:image/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==
data:image/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==

You can do this with just requests and re. It is only the co-ordinates part of the src which are the location based variable.
import requests, re
p = re.compile(r'"coords":"(.*?)"')
r = requests.get('https://www.tripadvisor.dk/Restaurant_Review-g189541-d15804886-Reviews-The_Pescatarian-Copenhagen_Zealand.html')
coords = p.findall(r.text)[1]
src = f'https://trip-raster.citymaps.io/staticmap?scale=1&zoom=15&size=347x137&language=da&center={coords}&markers=icon:http://c1.tacdn.com//img2/maps/icons/component_map_pins_v1/R_Pin_Small.png|{coords}'
print(src)
print(coords)

Selenium is a workaround i tested it and works liek a charm. Here you are:
from selenium import webdriver
driver = webdriver.Chrome('chromedriver.exe')
driver.get("https://www.tripadvisor.dk/Restaurant_Review-g189541-d15804886-Reviews-The_Pescatarian-Copenhagen_Zealand.html")
links = driver.find_elements_by_xpath("//*[#src]")
urls = []
for link in links:
url = link.get_attribute('src')
if '|' in url:
urls.append(url.split('|')[1]) # saves in a list only the numbers you want i.e. 55.68799,12.596316
print(url)
print(urls)
Result of above
['55.68799,12.596316']
If you haven't used selenium before here you can find a webdriver https://chromedriver.storage.googleapis.com/index.html?path=2.46/
or here
https://sites.google.com/a/chromium.org/chromedriver/downloads

How to save graph / image from CGI website in python?

http://www.wunderground.com/history/airport/KMDW/2014/11/17/MonthlyHistory.html?req_city=NA&req_state=NA&req_statename=NA
On the link above, I am trying to save the "Monthly Weather History Graph" in a python script. I have tried everything I can think of using BeautifulSoup and urrlib.
What I have been able to do is get to the point below, which I can extract, but I can not figure out how to save that graph as an image/HTML/PDF/anything. I am really not familiar with CGI, so any guidance here is much appreciated.
div id="history-graph-image"
img src="/cgi-bin/histGraphAll?day=17&year=2014&month=11&ID=KMDW&type=1&width=614**" alt="Monthly Weather History Graph" /

Get the page with requests, parse the HTML with BeautifulSoup, find the img tag inside div with id="history-graph-image" and get the src attribute value:
from urlparse import urljoin
from bs4 import BeautifulSoup
import requests
base_url = 'http://www.wunderground.com'
url = 'http://www.wunderground.com/history/airport/KMDW/2014/11/17/MonthlyHistory.html?req_city=NA&req_state=NA&req_statename=NA'
response = requests.get(url)
soup = BeautifulSoup(response.content)
image_relative_url = soup.find('div', id='history-graph-image').img.get('src')
image_url = urljoin(base_url, image_relative_url)
print image_url
Prints:
http://www.wunderground.com/cgi-bin/histGraphAll?day=17&year=2014&month=11&ID=KMDW&type=1&width=614
Then, download the file with urllib.urlretrieve():
import urllib
urllib.urlretrieve(image_url, "image.gif")