I'm new to web scraping with python, so I don't know if I'm doing this right.
I'm using a script that calls BeautifulSoup to parse the URLs from the first 10 pages of a google search. Tested with stackoverflow.com, worked just fine out-of-the-box. I tested with another site a few times, trying to see if the script was really working with higher google page requests, then it 503'd on me. I switched to another URL to test and worked for a couple, low-page requests, then also 503'd. Now every URL I pass to it is 503'ing. Any suggestions?
import sys # Used to add the BeautifulSoup folder the import path
import urllib2 # Used to read the html document
if __name__ == "__main__":
### Import Beautiful Soup
### Here, I have the BeautifulSoup folder in the level of this Python script
### So I need to tell Python where to look.
sys.path.append("./BeautifulSoup")
from BeautifulSoup import BeautifulSoup
### Create opener with Google-friendly user agent
opener = urllib2.build_opener()
opener.addheaders = [('User-agent', 'Mozilla/5.0')]
### Open page & generate soup
### the "start" variable will be used to iterate through 10 pages.
for start in range(0,10):
url = "http://www.google.com/search?q=site:stackoverflow.com&start=" + str(start*10)
page = opener.open(url)
soup = BeautifulSoup(page)
### Parse and find
### Looks like google contains URLs in <cite> tags.
### So for each cite tag on each page (10), print its contents (url)
for cite in soup.findAll('cite'):
print cite.text
Automated querying is not permitted by Google Terms of Service.
See this article for information:
Unusual traffic from your computer
and also Google Terms of service
As Ettore said, scraping the search results is against our ToS. However check out the WebSearch api, specifically the bottom section of the documentation which should give you a hint about how to access the API from non-javascipt environments.
Related
Novice web scraper here:
I am trying to scrape the name and address from this website https://propertyinfo.knoxcountytn.gov/Datalets/Datalet.aspx?sIndex=1&idx=1. I have attempted the following code which only returns 'None' or an empty array if I replace find() with find_all(). I would like it to return the html of this particular section so I can extract the text and later add it to a csv file. If the link doesn't work, or take to you where I'm working, simply go to the knox county tn website > property search > select a property.
Much appreciation in advance!
from splinter import Browser
import pandas as pd
from bs4 import BeautifulSoup as soup
import requests
from webdriver_manager.chrome import ChromeDriverManager
owner_soup = soup(html, 'html.parser')
owner_elem = owner_soup.find('td', class_='DataletData')
owner_elem
OR
# this being the tag and class of the whole section where the info is located
owner_soup = soup(html, 'html.parser')
owner_elem = owner_soup.find_all('div', class_='datalet_div_2')
owner_elem
OR when I try:
browser.find_by_css('td.DataletData')[15]
it returns:
<splinter.driver.webdriver.WebDriverElement at 0x11a763160>
and I can't pull the html contents from that element.
There's a few issues I see, but it could be that you didn't include your code as you actually have it.
Splinter works on its own to get page data by letting you control a browser. You don't need BeautifulSoup or requests if you're using splinter. You use requests if you want the raw response without running any of the things that browsers do for you automatically.
One of these automatic things is redirects. The link you provided does not provide the HTML that you are seeing. This link just has a response header that redirects you to https://propertyinfo.knoxcountytn.gov/, which redirects you again to https://propertyinfo.knoxcountytn.gov/search/commonsearch.aspx?mode=realprop, which redirects again to https://propertyinfo.knoxcountytn.gov/Search/Disclaimer.aspx?FromUrl=../search/commonsearch.aspx?mode=realprop
On this page you have to hit the 'agree' button to get redirected to https://propertyinfo.knoxcountytn.gov/search/commonsearch.aspx?mode=realprop, this time with these cookies set:
Cookie: ASP.NET_SessionId=phom3bvodsgfz2etah1wwwjk; DISCLAIMER=1
I'm assuming the session id is autogenerated, and the Disclaimer value just needs to be '1' for the server to know you agreed to their terms.
So you really have to study a page and understand what's going on to know how to do it on your own using just the requests and beautifulsoup libraries. Besides the redirects I mentioned, you still have to figure out what network request gives you that session id to manually add it to the cookie header you send on all future requests. You can avoid doing some requests, and so this way is a lot faster, but you do need to be able to follow along in the developer tools 'network' tab.
Postman is a good tool to help you set up requests yourself and see their result. Then you can bring all the set up from there into your code.
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 want to scrap data in the <span/> attribute for a given website using BeautifulSoup. You can see at the screenshot where it locates. However, the code that I'm using is just returning an empty list. I can't find the data in the list that I want. What am I doing wrong?
from bs4 import BeautifulSoup
from urllib import request
url = "http://144.122.167.229"
opener = urllib.request.build_opener()
opener.addheaders = [('User-agent', 'Mozilla/5.0')]
data = opener.open(url).read()
soup = BeautifulSoup(data, 'html.parser')
your_data = list()
for line in soup.findAll('span', attrs={'id': 'mc1_legend_value'}):
your_data.append(line.text)
for line in soup.findAll('span'):
your_data.append(line.text)
ScreenShot : https://imgur.com/a/z0vNh
Thank you.
The dashboard from the screenshot looks to me like something javascript would generate. If you can't find the tag in the page source, that means it was later added by some javascript code or your browser tried to fix some html which it considered broken or out of place.
Keep in mind that right now you're sending a request to a server and it serves you the plain html back. A browser would parse the html and execute any javascript code if it finds any. In your case, beautiful soup or urllib doesn't execute any javascript code. urllib fetches the html and beautiful soup makes it easier to parse and extract relevant information.
If you want to get the value from that tag, I recommend using a headless browser to render your page and just after that parse it's html through beautiful soup or any other parser.
Give a try to selenium: http://selenium-python.readthedocs.io/.
You can control your own browser programmatically. You can make it request the page for you, render it, save the new html in a variable, parse it using beautifoul soup and extract the values you're interested in. I believe that it already has it's own parser implemented which you can use directly to search for that tag.
Or maybe even scrapinghub's splash: https://github.com/scrapinghub/splash
If the dashboard communicates with a server in real-time and that value is continuously received from the server, you could take a look at what requests are sent to the server in order to get that value. Take a look in developer console under the networks tab. Press F12 to open the developer console and click on Network. Refresh the page and you should get all the request send to the server along with the responses. Requests sent by the javascript are usually XMLHttpRequests. Click on XHR in the Network tab to filter out any other requests. (These are instructions for Google Chrome. Firefox might differ a bit).
I have written a script that scrapes a URL. It works fine on Linux OS. But i am getting http 503 error when running on Windows 7. The URL has some issue.
I am using python 2.7.11 .
Please help.
Below is the script:
import sys # Used to add the BeautifulSoup folder the import path
import urllib2 # Used to read the html document
if __name__ == "__main__":
### Import Beautiful Soup
### Here, I have the BeautifulSoup folder in the level of this Python script
### So I need to tell Python where to look.
sys.path.append("./BeautifulSoup")
from bs4 import BeautifulSoup
### Create opener with Google-friendly user agent
opener = urllib2.build_opener()
opener.addheaders = [('User-agent', 'Mozilla/5.0')]
### Open page & generate soup
### the "start" variable will be used to iterate through 10 pages.
for start in range(0,1000):
url = "http://www.google.com/search?q=site:theknot.com/us/&start=" + str(start*10)
page = opener.open(url)
soup = BeautifulSoup(page)
### Parse and find
### Looks like google contains URLs in <cite> tags.
### So for each cite tag on each page (10), print its contents (url)
file = open("parseddata.txt", "wb")
for cite in soup.findAll('cite'):
print cite.text
file.write(cite.text+"\n")
# file.flush()
# file.close()
In case you run it in windows 7, the cmd throws http503 error stating the issue is with url.
The URL works fine in Linux OS. In case URL is actually wrong please suggest the alternatives.
Apparently with Python 2.7.2 on Windows, any time you send a custom User-agent header, urllib2 doesn't send that header. (source: https://stackoverflow.com/a/8994498/6479294).
So you might want to consider using requests instead of urllib2 in Windows:
import requests
# ...
page = requests.get(url)
soup = BeautifulSoup(page.text)
# etc...
EDIT: Also a very good point to be made is that Google may be blocking your IP - they don't really like bots making 100 odd requests sequentially.
This is probably a very simple task, but I cannot find any help. I have a website that takes the form www.xyz.com/somestuff/ID. I have a list of the IDs I need information from. I was hoping to have a simple script to go one the site and download the (complete) web page for each ID in a simple form ID_whatever_the_default_save_name_is in a specific folder.
Can I run a simple python script to do this for me? I can do it by hand, it is only 75 different pages, but I was hoping to use this to learn how to do things like this in the future.
Mechanize is a great package for crawling the web with python. A simple example for your issue would be:
import mechanize
br = mechanize.Browser()
response = br.open("www.xyz.com/somestuff/ID")
print response
This simply grabs your url and prints the response from the server.
This can be done simply in python using the urllib module. Here is a simple example in Python 3:
import urllib.request
url = 'www.xyz.com/somestuff/ID'
req = urllib.request.Request(url)
page = urllib.request.urlopen(req)
src = page.readall()
print(src)
For more info on the urllib module -> http://docs.python.org/3.3/library/urllib.html
Do you want just the html code for the website? If so, just create a url variable with the host site and add the page number as you go. I'll do this for an example with http://www.notalwaysright.com
import urllib.request
url = "http://www.notalwaysright.com/page/"
for x in range(1, 71):
newurl = url + x
response = urllib.request.urlopen(newurl)
with open("Page/" + x, "a") as p:
p.writelines(reponse.read())