How can we print the URL's which appears in the Google's results in the python in version 2.7?
Here is the code,
domain = sys.argv[1];
print domain;
test = []
test.append("aa")
mainURL = "http://google.com/?q=";
finalurl = mainURL + test[0]
req = requests.get(finalurl)
How can I print the URL's after firing up the request.get(finalurl)?
Normally it would be:
r = requests.get("url");
from bs4 import BeautifulSoup
soup = BeautifulSoup(r.text, 'html.parser')
soup.find_all("cite")
You can look at the website's source code (ctrl+U, or dev-tools/Inspector and find the tags/class/id where your data is. I see e.g.: <cite class="iUh30">link...)
Now the problem is here, if you look at, the output of:
print(soup.prettify())
Then you see, they do not just give you back plain-text, but generate it with JS.
You could have a look at other search engines responses
Dig deeper into Js code execution from python (maybe other libs etc?)
Please, also think about, that some would not consider parsing out data from other websites programatically and in big amount without permission legit. But I have no idea what google's policy is regarding this.
requests advanced usage
bs4
=======
EDIT:
#Rahul's link is the fastest solution to OP.
https://www.geeksforgeeks.org/performing-google-search-using-python-code/
Related
I am trying to extract information about prices of flight tickets with a python script. Please take a look at the picture:
I would like to parse all the prices (such as "121" at the bottom of the tree). I have constructed a simple script and my problem is that I am not sure how to get the right parts from the code behind page's "inspect element". My code is below:
import urllib3
from bs4 import BeautifulSoup as BS
http = urllib3.PoolManager()
ULR = "https://greatescape.co/?datesType=oneway&dateRangeType=exact&departDate=2019-08-19&origin=EAP&originType=city&continent=europe&flightType=3&city=WAW"
response = http.request('GET', URL)
soup = BS(response.data, "html.parser")
body = soup.find('body')
__next = body.find('div', {'id':'__next'})
ui_container = __next.find('div', {'class':'ui-container'})
bottom_container_root = ui_container.find('div', {'class':'bottom-container-root'})
print(bottom_container_root)
The problem is that I am stuck at the level of ui-container. bottom-container-root is an empty variable, despite it is a direct child under ui-container. Could someone please let me know how to parse this tree properly?
I have no experience in web scraping, but as it happens it is one step in a bigger workflow I am building.
.find_next_siblings and .next_element can be useful in navigating through containers.
Here is some example usage below.
from bs4 import BeautifulSoup
html = open("small.html").read()
soup = BeautifulSoup(html)
print soup.head.next_element
print soup.head.next_element.next_element
I want to open a website to download resume from it, but following code tries to get to absolute path instead of just url:
import webbrowser
soup = BeautifulSoup(webbrowser.open('www.indeed.com/r/Prabhanshu-Pandit/dee64d1418e20069?sp=0'),"lxml")
generates the following error:
gvfs-open: /home/utkarsh/Documents/Extract_Resume/www.indeed.com/r/Prabhanshu-
Pandit/dee64d1418e20069?sp=0:
error opening location: Error when getting information for file
'/home/utkarsh/Documents/Extract_Resume/www.indeed.com/r/Prabhanshu-
Pandit/dee64d1418e20069?sp=0': No such file or directory
Clearly it is taking the home address and trying to search that on web which will not be present. What am I doing wrong here? Thanks in advance
I suppose you are confusing the usage of Beautiful Soup and webbrowser together. Webbrowser it is not needed to access the page.
From Documentation
Beautiful Soup provides a few simple methods and Pythonic idioms for
navigating, searching, and modifying a parse tree: a toolkit for
dissecting a document and extracting what you need. It doesn't take
much code to write an application
Adapting the tutorial example to your task to print the resume in output
from bs4 import BeautifulSoup
import requests
url = "www.indeed.com/r/Prabhanshu-Pandit/dee64d1418e20069?sp=0"
r = requests.get("http://" +url)
data = r.text
soup = BeautifulSoup(data, "html.parser")
print soup.find("div", {"id": "resume"})
I want to build small tool to help a family member download podcasts off a site.
In order to get the links to the files I first need to filter them out (with bs4 + python3).
The files are on this website (Estonian): Download Page "Laadi alla" = "Download"
So far my code is as follows:
(most of it is from examples on stackoverflow)
from bs4 import BeautifulSoup
import urllib.request
import re
url = urllib.request.urlopen("http://vikerraadio.err.ee/listing/mystiline_venemaa#?page=1&pagesize=902&phrase=&from=&to=&path=mystiline_venemaa&showAll")
content = url.read()
soup = BeautifulSoup(content, "lxml")
links = [a['href'] for a in soup.find_all('a',href=re.compile('http.*\.mp3'))]
print ("Links:", links)
Unfortunately I always get only two results.
Output:
Links: ['http://heli.err.ee/helid/exp/ERR_raadiouudised.mp3', 'http://heli.err.ee/helid/exp/ERR_raadiouudised.mp3']
These are not the ones I want.
My best guess is that the page has somewhat broken html and bs4 / the parser is not able to find anything else.
I've tried different parsers with resulting in no change.
Maybe I'm doing something else wrong too.
My goal is to have the individual links in a list for example.
I'll filter out any duplicates / unwanted entries later myself.
Just a quick note, just in case: This is a public radio and all the content is legally hosted.
My new code is:
for link in soup.find_all('d2p1:DownloadUrl'):
print(link.text)
I am very unsure if the tag is selected correctly.
None of the examples listed in this question are actually working. See the answer below for working code.
Please be aware that the listings from the page are interfaced through an API. So instead of requesting the HTML page, I suggest you to request the API link which has 200 .mp3 links.
Please follow the below steps:
Request the API link, not the HTML page link
Check the response, it's a JSON. So extract the fields that are of your need
Help your Family, All Time :)
Solution
import requests, json
from bs4 import BeautifulSoup
myurl = 'http://vikerraadio.err.ee/api/listing/bypath?path=mystiline_venemaa&page=1&pagesize=200&phrase=&from=&to=&showAll=false'
r = requests.get(myurl)
abc = json.loads(r.text)
all_mp3 = {}
for lstngs in abc['ListItems']:
for asd in lstngs['Podcasts']:
all_mp3[asd['DownloadUrl']] = lstngs['Header']
all_mp3
all_mp3 is what you need. all_mp3 is a dictionary with download urls as keys and mp3 names as the values.
i want to be able to pull all urls from the following webpage using python https://yeezysupply.com/pages/all i tried using some other suggestions i found but they didn't seem to work with this particular website. i would end up not finding any urls at all.
import urllib
import lxml.html
connection = urllib.urlopen('https://yeezysupply.com/pages/all')
dom = lxml.html.fromstring(connection.read())
for link in dom.xpath('//a/#href'):
print link
perhaps it would be useful for you to make use of modules specifically designed for this. heres a quick and dirty script that gets the relative links on the page
#!/usr/bin/python3
import requests, bs4
res = requests.get('https://yeezysupply.com/pages/all')
soup = bs4.BeautifulSoup(res.text,'html.parser')
links = soup.find_all('a')
for link in links:
print(link.attrs['href'])
it generates output like this:
/pages/jewelry
/pages/clothing
/pages/footwear
/pages/all
/cart
/products/womens-boucle-dress-bleach/?back=%2Fpages%2Fall
/products/double-sleeve-sweatshirt-bleach/?back=%2Fpages%2Fall
/products/boxy-fit-zip-up-hoodie-light-sand/?back=%2Fpages%2Fall
/products/womens-boucle-skirt-cream/?back=%2Fpages%2Fall
etc...
is this what you are looking for? requests and beautiful soup are amazing tools for scraping.
There are no links in the page source; they are inserted using Javascript after the page is loaded int the browser.
I am using python 2.7 and version 4.5.1 of Beautiful Soup
I'm at my wits end trying to make this very simple script to work. My goal is to to get the information on the online availability status of the NES console from Best Buy's website by parsing the html for the product's page and extracting the information in
<div class="status online-availability-status"> Sold out online </div>
This is my first time using the Beautiful Soup module so forgive me if I have missed something obvious. Here is the script I wrote to try to get the information above:
import requests
from bs4 import BeautifulSoup
page = requests.get('http://www.bestbuy.ca/en-CA/product/nintendo-nintendo-entertainment-system-nes-classic-edition-console-clvsnesa/10488665.aspx?path=922de2a5ceb066b0f058cc567ad3d547en02')
soup = BeautifulSoup(page.content, 'html.parser')
avail = soup.findAll('div', {"class": "status online-availability-status"})
But then I just get an empty list for avail. Any idea why?
Any help is greatly appreciated.
As the comments above suggest, it seems that you are looking for a tag which is generated client side by JavaScript; it shows up using 'inspect' on the loaded page, but not when viewing the page source, which is what the call to requests is pulling back. You might try using dryscrape (which you may need to install with pip install dryscrape).
import dryscrape
from bs4 import BeautifulSoup
session = dryscrape.Session()
url = 'http://www.bestbuy.ca/en-CA/product/nintendo-nintendo-entertainment-system-nes-classic-edition-console-clvsnesa/10488665.aspx?path=922de2a5ceb066b0f058cc567ad3d547en02'
session.visit(url)
response = session.body()
soup = BeautifulSoup(response)
avail = soup.findAll('div', {"class": "status online-availability-status"})
This was the most popular solution in a question relating to scraping dynamically generated content:
Web-scraping JavaScript page with Python
If you try printing soup you'll see it probably returns something like Access Denied. This is because Best Buy requires an allowable User-Agent to be making the GET request. As you do not have a User-Agent specified in the Header, it is not returning anything.
Here is a link to generate a User Agent
How to use Python requests to fake a browser visit a.k.a and generate User Agent?
or you could figure out your user agent generated when you are viewing the webpage in your own browser
https://developer.mozilla.org/en-US/docs/Web/HTTP/Headers/User-Agent
Availability is loaded in JSON. You don't even need to parse HTML for that:
import urllib
import simplejson
sku = 1048865 # look at the URL of the web page, it is <blablah>//10488665.aspx
# chnage locations to get the right store
response = urllib.urlopen('http://api.bestbuy.ca/availability/products?callback=apiAvailability&accept-language=en&skus=%s&accept=application%2Fvnd.bestbuy.standardproduct.v1%2Bjson&postalCode=M5G2C3&locations=977%7C203%7C931%7C62%7C617&maxlos=3'%sku)
availability = simplejson.loads(response.read())
print availability[0]['shipping']['status']