I'm trying to catch dynamic content from a webpage. The data is displayed dynamically on the webpage after loading the content.
On one webpage the response in the console is json formatted and html for the second one.
I've tried to work with scrappy and urllib3 but did not manage to catch something else then the static data from the webpage itself.
Here is what I've tried to use with scrappy.
class spider(scrapy.Spider):
name = 'myspider'
start_urls = [url]
def parse(self, response):
yield scrapy.FormRequest('myurl',
callback=self.write_vente,
headers=headers,
meta={'proxy': 'https://' + str(proxy)})
def write_vente(self, response):
filename = 'vente.html'
with open(filename, 'wb') as f:
f.write(response.body)
If you know any solutions or other libraries/framework to use or even other programming language that allows me to do so
Thanks
The most common used tool to scrape data from dynamic websites is Selenium WebDriver. Which also has good support for Python, can be used headless. Also it has loads of articles if you search for it incombination with scraping.
Scrappy does have some support for pre-loading dynamic content or using Selenium in combination with Scrappy, see: https://docs.scrapy.org/en/latest/topics/dynamic-content.html#topics-javascript-rendering
Related
I want to scrape website link from "https://www.theknot.com/marketplace/bayside-bowl-portland-me-1031451", and I have used proper code of xpath i.e. response.xpath("//a[#title='website']/#href").get(), but it shows null result while scraping.
Some websites (in fact, many) use Javascript to generate content. That's why you need always check a source HTML code.
For this page you'll find that all information you need is inside a script tag with this text window.__INITIAL_STATE__ = . You need to get this text and next you can use json module to parse it:
import scrapy
import json
class TheknotSpider(scrapy.Spider):
name = 'theknot'
start_urls = ['https://www.theknot.com/marketplace/bayside-bowl-portland-me-1031451']
def parse(self, response):
initial_state_raw = response.xpath('//script[contains(., "window.__INITIAL_STATE__ = ")]/text()').re_first(r'window\.__INITIAL_STATE__ = (\{.+?)$')
# with open('Samples/TheKnot.json', 'w', encoding='utf-8') as f:
# f.write(initial_state_raw)
initial_state = json.loads(initial_state_raw)
website = initial_state['vendor']['vendor']['displayWebsiteUrl']
print(website)
I am working on a web scraping project and want to get a list of products from Dell's website. I found this link (https://www.dell.com/support/home/us/en/04/products/) which pulls up a box with a list of product categories (really just redirect urls. If it doesn't come up for you click the button which says "Browse all products"). I tried using Python Requests to GET the page and save the text to a file to parse through, but the response doesn't contain any of the categories/redirect urls. My code is as basic as it gets:
import requests
url = "https://www.dell.com/support/home/us/en/04/products/"
page = requests.get(url)
with open("laptops.txt", "w", encoding="utf-8") as outf:
outf.write(page.text)
outf.close()
Is there a way to get these redirect urls? I am essentially trying to make my own site map of their products so that I can scrape the details of each one. Thanks
This page uses JavaScript to get and display these links - but requests/urllib and BeautifulSoup/lxml can't run JavaScript.
Using DevTools in Firefox/Chrome (tab: Network) I found it reads it from url
https://www.dell.com/support/components/productselector/allproducts?category=all-products/esuprt_&country=pl&language=pl®ion=emea&segment=bsd&customerset=plbsd1&openmodal=true&_=1589265310743
so I use it to get links.
You may have to to change country=pl&language=pl in url to get it in different language.
import requests
from bs4 import BeautifulSoup as BS
url = "https://www.dell.com/support/components/productselector/allproducts?category=all-products/esuprt_&country=pl&language=pl®ion=emea&segment=bsd&customerset=plbsd1&openmodal=true&_=1589265310743"
response = requests.get(url)
soup = BS(response.text, 'html.parser')
all_items = soup.find_all('a')
for item in all_items:
print(item.text, item['href'])
BTW: Other method is it use Selenium to control real web browser which can run JavaScript.
try using selenium chrome driver it helps for handling dynamic data on website and also features like clicking buttons, handling page refresh etc.
Beginner guide to web scraping
The original code is here : https://github.com/amitabhadey/Web-Scraping-Images-using-Python-via-BeautifulSoup-/blob/master/code.py
So i am trying to adapt a Python script to collect pictures from a website to get better at web scraping.
I tried to get images from "https://500px.com/editors"
The first error was
The code that caused this warning is on line 12 of the file/Bureau/scrapper.py. To get rid of this warning, pass the additional argument
'features="lxml"' to the BeautifulSoup constructor.
So I did :
soup = BeautifulSoup(plain_text, features="lxml")
I also adapted the class to reflect the tag in 500px.
But now the script stopped running and nothing happened.
In the end it looks like this :
import requests
from bs4 import BeautifulSoup
import urllib.request
import random
url = "https://500px.com/editors"
source_code = requests.get(url)
plain_text = source_code.text
soup = BeautifulSoup(plain_text, features="lxml")
for link in soup.find_all("a",{"class":"photo_link "}):
href = link.get('href')
print(href)
img_name = random.randrange(1,500)
full_name = str(img_name) + ".jpg"
urllib.request.urlretrieve(href, full_name)
print("loop break")
What did I do wrong?
Actually the website is loaded via JavaScript using XHR request to the following API
So you can reach it directly via API.
Note that you can increase parameter rpp=50 to any number as you want for getting more than 50 result.
import requests
r = requests.get("https://api.500px.com/v1/photos?rpp=50&feature=editors&image_size%5B%5D=1&image_size%5B%5D=2&image_size%5B%5D=32&image_size%5B%5D=31&image_size%5B%5D=33&image_size%5B%5D=34&image_size%5B%5D=35&image_size%5B%5D=36&image_size%5B%5D=2048&image_size%5B%5D=4&image_size%5B%5D=14&sort=&include_states=true&include_licensing=true&formats=jpeg%2Clytro&only=&exclude=&personalized_categories=&page=1&rpp=50").json()
for item in r['photos']:
print(item['url'])
also you can access the image url itself in order to write it directly!
import requests
r = requests.get("https://api.500px.com/v1/photos?rpp=50&feature=editors&image_size%5B%5D=1&image_size%5B%5D=2&image_size%5B%5D=32&image_size%5B%5D=31&image_size%5B%5D=33&image_size%5B%5D=34&image_size%5B%5D=35&image_size%5B%5D=36&image_size%5B%5D=2048&image_size%5B%5D=4&image_size%5B%5D=14&sort=&include_states=true&include_licensing=true&formats=jpeg%2Clytro&only=&exclude=&personalized_categories=&page=1&rpp=50").json()
for item in r['photos']:
print(item['image_url'][-1])
Note that image_url key hold different img size. so you can choose your preferred one and save it. here I've taken the big one.
Saving directly:
import requests
with requests.Session() as req:
r = req.get("https://api.500px.com/v1/photos?rpp=50&feature=editors&image_size%5B%5D=1&image_size%5B%5D=2&image_size%5B%5D=32&image_size%5B%5D=31&image_size%5B%5D=33&image_size%5B%5D=34&image_size%5B%5D=35&image_size%5B%5D=36&image_size%5B%5D=2048&image_size%5B%5D=4&image_size%5B%5D=14&sort=&include_states=true&include_licensing=true&formats=jpeg%2Clytro&only=&exclude=&personalized_categories=&page=1&rpp=50").json()
result = []
for item in r['photos']:
print(f"Downloading {item['name']}")
save = req.get(item['image_url'][-1])
name = save.headers.get("Content-Disposition")[9:]
with open(name, 'wb') as f:
f.write(save.content)
Looking at the page you're trying to scrape I noticed something. The data doesn't appear to load until a few moments after the page finishes loading. This tells me that they're using a JS framework to load the images after page load.
Your scraper will not work with this page due to the fact that it does not run JS on the pages it's pulling. Running your script and printing out what plain_text contains proves this:
<a class='photo_link {{#if hasDetailsTooltip}}px_tooltip{{/if}}' href='{{photoUrl}}'>
If you look at the href attribute on that tag you'll see it's actually a templating tag used by JS UI frameworks.
Your options now are to either see what APIs they're calling to get this data (check the inspector in your web browser for network calls, if you're lucky they may not require authentication) or to use a tool that runs JS on pages. One tool I've seen recommended for this is selenium, though I've never used it so I'm not fully aware of its capabilities; I imagine the tooling around this would drastically increase the complexity of what you're trying to do.
I am building a crawl.spider to scrape statutory law data from the following website (https://www.azleg.gov/viewdocument/?docName=https://www.azleg.gov/ars/1/00101.htm). I am aiming to extract the statute text, which is contained in the following XPath [//div[#class = 'first']/p/text()]. This path should provide the statute text.
All of my scrapy requests are yielding incomplete html responses, such that when I search for the relevant xpath queries, it yields an empty list. However, when I use the requests library, the html downloads correctly.
Using XPath tester online, I've verified that my xpath queries should produce the desired content. Using scrapy shell, I've viewed the response object from scrapy in my browser - and it looks just like it does when I'm browsing natively. I've tried enabling middleware for both BeautifulSoup and Selenium, but neither has appeared to work.
Here's my crawl spider
class AZspider(CrawlSpider):
name = "arizona"
start_urls = [
"https://www.azleg.gov/viewdocument/?docName=https://www.azleg.gov/ars/1/00101.htm",
]
rule = (Rule(LinkExtractor(restrict_xpaths="//div[#class = 'article']"), callback="parse_stats_az", follow=True),)
def parse_stats_az(self, response):
statutes = response.xpath("//div[#class = 'first']/p")
yield{
"statutes":statutes
}
And here's the code that succsessfuly generated the correct response object
az_leg = requests.get("https://www.azleg.gov/viewdocument/?docName=https://www.azleg.gov/ars/1/00101.htm")
I have the following code for a web crawler in Python 3:
import requests
from bs4 import BeautifulSoup
import re
def get_links(link):
return_links = []
r = requests.get(link)
soup = BeautifulSoup(r.content, "lxml")
if r.status_code != 200:
print("Error. Something is wrong here")
else:
for link in soup.findAll('a', attrs={'href': re.compile("^http")}):
return_links.append(link.get('href')))
def recursive_search(links)
for i in links:
links.append(get_links(i))
recursive_search(links)
recursive_search(get_links("https://www.brandonskerritt.github.io"))
The code basically gets all the links off of my GitHub pages website, and then it gets all the links off of those links, and so on until the end of time or an error occurs.
I want to recreate this code in Scrapy so it can obey robots.txt and be a better web crawler overall. I've researched online and I can only find tutorials / guides / stackoverflow / quora / blog posts about how to scrape a specific domain (allowed_domains=["google.com"], for example). I do not want to do this. I want to create code that will scrape all websites recursively.
This isn't much of a problem but all the blog posts etc only show how to get the links from a specific website (for example, it might be that he links are in list tags). The code I have above works for all anchor tags, regardless of what website it's being run on.
I do not want to use this in the wild, I need it for demonstration purposes so I'm not going to suddenly annoy everyone with excessive web crawling.
Any help will be appreciated!
There is an entire section of scrapy guide dedicated to broad crawls. I suggest you to fine-grain your settings for doing this succesfully.
For recreating the behaviour you need in scrapy, you must
set your start url in your page.
write a parse function that follow all links and recursively call itself, adding to a spider variable the requested urls
An untested example (that can be, of course, refined):
class AllSpider(scrapy.Spider):
name = 'all'
start_urls = ['https://yourgithub.com']
def __init__(self):
self.links=[]
def parse(self, response):
self.links.append(response.url)
for href in response.css('a::attr(href)'):
yield response.follow(href, self.parse)
If you want to allow crawling of all domains, simply don't specify allowed_domains, and use a LinkExtractor which extracts all links.
A simple spider that follows all links:
class FollowAllSpider(CrawlSpider):
name = 'follow_all'
start_urls = ['https://example.com']
rules = [Rule(LinkExtractor(), callback='parse_item', follow=True)]
def parse_item(self, response):
pass