I am trying to extract the source code of the html page. It was working fine before. but now the source web server wanted more evidence that I am NOT a bot. This is the error: your IP is blocked. My IP is NOT blocked for sure as I can still open the page manually via any browser. Do I need to change any parameters before making the request. Thanks.
headers = {"User-Agent": "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/70.0.3538.77 Safari/537.36"}
req = requests.get(url, headers=headers)
url_content = req.content
url_content = url_content.replace(b'data-imgid', b'\ndata-imgid')
output_file = open('downloaded.txt', 'wb')
output_file.write(url_content)
output_file.close()
Related
I am trying to scrape the following web page
https://jamanetwork.com/journals/jamaneurology/article-abstract/2696970
but getting an error.
url ='https://jamanetwork.com/journals/jamaneurology/article-abstract/2696970'
result = requests.get(url)
soup = BeautifulSoup(result.content, 'html.parser')
print(soup.prettify())
Result:
403 Forbidden Request forbidden by
administrative rules.
You can access the web page with no credentials, so not sure why I get 'Request forbidden' error while scraping.
As mentioned you should add a user-agent to your request:
headers = {'user-agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/96.0.4664.110 Safari/537.36'}
You can check your own headers, send by the browser via opening dev tools and take a look under network section. Read more about user-agent.
Example
headers = {'user-agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/96.0.4664.110 Safari/537.36'}
url ='https://jamanetwork.com/journals/jamaneurology/article-abstract/2696970'
result = requests.get(url,headers=headers)
soup = BeautifulSoup(result.content, 'html.parser')
print(soup.prettify())
I try to learn Web Scraping using python. 1st I try to scrape from an amazon page. I try to find out the Best Sellers in Women's Fashion Sneakers.
My code:
no_pages = 2
def get_data(pageNo):
headers = {"User-Agent":"Mozilla/5.0 (Windows NT 10.0; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/88.0.4324.182 Safari/537.36", "Accept-Encoding":"gzip, deflate", "Accept":"text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8", "DNT":"1","Connection":"close", "Upgrade-Insecure-Requests":"1"}
r = requests.get('https://www.amazon.com/Best-Sellers-Womens-Fashion-Sneakers/zgbs/fashion/679394011'+str(pageNo)+'?ie=UTF8&pg='+str(pageNo), headers=headers)#, proxies=proxies)
content = r.content
soup = BeautifulSoup(content)
print(soup)
I did not get any output for this portion. Then I find out this. I thought that web scraping is not possible for Amazon. Then I changed my source page to monster.com. But did not get the output.
no_pages = 2
def get_data(pageNo):
headers = {"User-Agent":"Mozilla/5.0 (Windows NT 10.0; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/88.0.4324.182 Safari/537.36", "Accept-Encoding":"gzip, deflate", "Accept":"text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8", "DNT":"1","Connection":"close", "Upgrade-Insecure-Requests":"1"}
r = requests.get('https://www.monster.com/jobs/search/?q=Software-developer&where=Texas-City__2C-TX'+str(pageNo)+'?ie=UTF8&pg='+str(pageNo), headers=headers)#, proxies=proxies)
content = r.content
soup = BeautifulSoup(content)
print(soup)
How could I solve this problem. Thank you.
It seems that your code runs correctly. However, you need to call the function in order to run it. For instance, after your function, write the following:
get_data(no_pages)
This will trigger the function to run.
import requests
headers = {'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/84.0.4147.135 Safari/537.36'}
url = 'https://www.nseindia.com/api/chart-databyindex?index=ACCEQN'
r = requests.get(url, headers=headers)
data = r.json()
print(data)
prices = data['grapthData']
print(prices)
It was working fine but now it showing error "Response [401]"
Well, it's all about the site's authentication requirements. It requires a certain level of authorization to access like this.
Im using the requests module in python to try and make a search on the following webiste http://musicpleer.audio/, however this website appears to be blocking me as it issues nothing but a 403 when i attempt to access it, im wondering how i can get around this, ive tried sending it the user agent of my web browser(chrome) and it still returns error 403. any suggestions on how i could get around this an example of downloading a song from the site would be very helpful. Thanks in advance
My code:
import requests, os
def funGetList:
start_path = 'C:/Users/Jordan/Music/' # current directory
list = []
for path,dirs,files in os.walk(start_path):
for filename in files:
temp = (os.path.join(path,filename))
tempLen = len(temp)
"print(tempLen)"
iterate = 0
list.append(temp[22:(len(temp))-4])
def funDownloadMP3:
for i in list:
print(i)
payload = {'searchQuery': 'meme', 'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/60.0.3112.113 Safari/537.36'}
url = 'http://musicpleer.audio/'
print(requests.post(url, data=payload))
Putting the User-Agent in the headers seems to work:
In []:
import requests
headers = {'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/60.0.3112.113 Safari/537.36'}
url = 'http://musicpleer.audio/'
r = requests.get('{}#!{}'.format(url, 'meme'), headers=headers)
r.status_code
Out[]:
200
Note: It looks like the search url is simple '#!<search-term>'
HTML 403 Forbidden error code.
The server might be expecting some more request headers like Host or Cookies etc.
You might want to use Postman to debug it with ease
for a university project I am currently trying to login to a website, and scrap a little detail (a list of news articles) from my user profile.
I am new to Python, but I did this before to some other website. My first two approaches deliver different HTTP errors. I have considered problems with the header my request is sending, however my understanding of this sites login process appears to be insufficient.
This is the login page: http://seekingalpha.com/account/login
My first approach looks like this:
import requests
with requests.Session() as c:
requestUrl ='http://seekingalpha.com/account/orthodox_login'
USERNAME = 'XXX'
PASSWORD = 'XXX'
userAgent = 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/49.0.2623.112 Safari/537.36'
login_data = {
"slugs[]":None,
"rt":None,
"user[url_source]":None,
"user[location_source]":"orthodox_login",
"user[email]":USERNAME,
"user[password]":PASSWORD
}
c.post(requestUrl, data=login_data, headers = {"referer": "http://seekingalpha.com/account/login", 'user-agent': userAgent})
page = c.get("http://seekingalpha.com/account/email_preferences")
print(page.content)
This results in "403 Forbidden"
My second approach looks like this:
from requests import Request, Session
requestUrl ='http://seekingalpha.com/account/orthodox_login'
USERNAME = 'XXX'
PASSWORD = 'XXX'
userAgent = 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/49.0.2623.112 Safari/537.36'
# c.get(requestUrl)
login_data = {
"slugs[]":None,
"rt":None,
"user[url_source]":None,
"user[location_source]":"orthodox_login",
"user[email]":USERNAME,
"user[password]":PASSWORD
}
headers = {
"accept":"text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,*/*;q=0.8",
"Accept-Language":"de-DE,de;q=0.8,en-US;q=0.6,en;q=0.4",
"origin":"http://seekingalpha.com",
"referer":"http://seekingalpha.com/account/login",
"Cache-Control":"max-age=0",
"Upgrade-Insecure-Requests":1,
"user-agent":"Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/49.0.2623.112 Safari/537.36"
}
s = Session()
req = Request('POST', requestUrl, data=login_data, headers=headers)
prepped = s.prepare_request(req)
prepped.body ="slugs%5B%5D=&rt=&user%5Burl_source%5D=&user%5Blocation_source%5D=orthodox_login&user%5Bemail%5D=XXX%40XXX.com&user%5Bpassword%5D=XXX"
resp = s.send(prepped)
print(resp.status_code)
In this approach I was trying to prepare the header exactly as my browser would do it. Sorry for redundancy. This results in HTTP error 400.
Does someone have an idea, what went wrong? Probably a lot.
Instead of spending a lot of energy on manually logging in and playing with Session, I suggest you just scrape the pages right away using your cookie.
When you log in, usually there is a cookie added to your request to identify your identity. Please see this for example:
Your code will be like this:
import requests
response = requests.get("www.example.com", cookies={
"c_user":"my_cookie_part",
"xs":"my_other_cookie_part"
})
print response.content