Why can't i scrape the particular Amazon music page with selenium python?
https://www.amazon.com/Prettymuch-EP-PRETTYMUCH/dp/B07CF6YXDP
The pure above mentioned link, not the link that one is directed to once clicking on it with the stack overflow tag.
This is the url.
def get_soup(url):
headers = {'User-Agent':
'Mozilla/5.0 (Windows NT 6.1) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/41.0.2228.0 Safari/537.36',
}
r = requests.get(url, headers=headers)
r.raise_for_status()
return BeautifulSoup(r.text, 'lxml')
url = input("Please enter an Amazon music url:")
soup = get_soup(url)
I get an error when requesting through it, why is that so?
Please enter an Amazon music url:https://www.amazon.com/Prettymuch-EP-
PRETTYMUCH/dp/B07CF6YXDP
Traceback (most recent call last):
File "D:/Pycharm (4)/selemin.py", line 4, in <module>
import amazon
File "D:Pycharm (4)amazon.py", line 63, in <module>
soup = get_soup(url)
File "D:Pycharm (4)amazon.py", line 12, in get_soup
r.raise_for_status()
File "C:UsersHPAppDataLocalProgramsPythonPython37-32libsite-
packagesrequestsmodels.py", line 940, in raise_for_status
raise HTTPError(http_error_msg, response=self)
requests.exceptions.HTTPError: 404 Client Error: Not Found for url:
https://www.amazon.com/Prettymuch-EP-PRETTYMUCH/dp/B07CF6YXDP%20
python beautifulsoup amazon
edited Nov 16 '18 at 3:16
Aqueous Carlos
299313
asked Nov 16 '18 at 3:06
Durian JaykinDurian Jaykin
126
add a comment |
https://www.amazon.com/Prettymuch-EP-PRETTYMUCH/dp/B07CF6YXDP
The pure above mentioned link, not the link that one is directed to once clicking on it with the stack overflow tag.
This is the url.
def get_soup(url):
headers = {'User-Agent':
'Mozilla/5.0 (Windows NT 6.1) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/41.0.2228.0 Safari/537.36',
}
r = requests.get(url, headers=headers)
r.raise_for_status()
return BeautifulSoup(r.text, 'lxml')
url = input("Please enter an Amazon music url:")
soup = get_soup(url)
I get an error when requesting through it, why is that so?
Please enter an Amazon music url:https://www.amazon.com/Prettymuch-EP-
PRETTYMUCH/dp/B07CF6YXDP
Traceback (most recent call last):
File "D:/Pycharm (4)/selemin.py", line 4, in <module>
import amazon
File "D:Pycharm (4)amazon.py", line 63, in <module>
soup = get_soup(url)
File "D:Pycharm (4)amazon.py", line 12, in get_soup
r.raise_for_status()
File "C:UsersHPAppDataLocalProgramsPythonPython37-32libsite-
packagesrequestsmodels.py", line 940, in raise_for_status
raise HTTPError(http_error_msg, response=self)
requests.exceptions.HTTPError: 404 Client Error: Not Found for url:
https://www.amazon.com/Prettymuch-EP-PRETTYMUCH/dp/B07CF6YXDP%20
python beautifulsoup amazon
edited Nov 16 '18 at 3:16
Aqueous Carlos
299313
asked Nov 16 '18 at 3:06
Durian JaykinDurian Jaykin
126
add a comment |
https://www.amazon.com/Prettymuch-EP-PRETTYMUCH/dp/B07CF6YXDP
The pure above mentioned link, not the link that one is directed to once clicking on it with the stack overflow tag.
This is the url.
def get_soup(url):
headers = {'User-Agent':
'Mozilla/5.0 (Windows NT 6.1) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/41.0.2228.0 Safari/537.36',
}
r = requests.get(url, headers=headers)
r.raise_for_status()
return BeautifulSoup(r.text, 'lxml')
url = input("Please enter an Amazon music url:")
soup = get_soup(url)
I get an error when requesting through it, why is that so?
Please enter an Amazon music url:https://www.amazon.com/Prettymuch-EP-
PRETTYMUCH/dp/B07CF6YXDP
Traceback (most recent call last):
File "D:/Pycharm (4)/selemin.py", line 4, in <module>
import amazon
File "D:Pycharm (4)amazon.py", line 63, in <module>
soup = get_soup(url)
File "D:Pycharm (4)amazon.py", line 12, in get_soup
r.raise_for_status()
File "C:UsersHPAppDataLocalProgramsPythonPython37-32libsite-
packagesrequestsmodels.py", line 940, in raise_for_status
raise HTTPError(http_error_msg, response=self)
requests.exceptions.HTTPError: 404 Client Error: Not Found for url:
https://www.amazon.com/Prettymuch-EP-PRETTYMUCH/dp/B07CF6YXDP%20
python beautifulsoup amazon
edited Nov 16 '18 at 3:16
Aqueous Carlos
299313
asked Nov 16 '18 at 3:06
Durian JaykinDurian Jaykin
126
https://www.amazon.com/Prettymuch-EP-PRETTYMUCH/dp/B07CF6YXDP
The pure above mentioned link, not the link that one is directed to once clicking on it with the stack overflow tag.
This is the url.
def get_soup(url):
headers = {'User-Agent':
'Mozilla/5.0 (Windows NT 6.1) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/41.0.2228.0 Safari/537.36',
}
r = requests.get(url, headers=headers)
r.raise_for_status()
return BeautifulSoup(r.text, 'lxml')
url = input("Please enter an Amazon music url:")
soup = get_soup(url)
I get an error when requesting through it, why is that so?
Please enter an Amazon music url:https://www.amazon.com/Prettymuch-EP-
PRETTYMUCH/dp/B07CF6YXDP
Traceback (most recent call last):
File "D:/Pycharm (4)/selemin.py", line 4, in <module>
import amazon
File "D:Pycharm (4)amazon.py", line 63, in <module>
soup = get_soup(url)
File "D:Pycharm (4)amazon.py", line 12, in get_soup
r.raise_for_status()
File "C:UsersHPAppDataLocalProgramsPythonPython37-32libsite-
packagesrequestsmodels.py", line 940, in raise_for_status
raise HTTPError(http_error_msg, response=self)
requests.exceptions.HTTPError: 404 Client Error: Not Found for url:
https://www.amazon.com/Prettymuch-EP-PRETTYMUCH/dp/B07CF6YXDP%20
python beautifulsoup amazon
python beautifulsoup amazon
edited Nov 16 '18 at 3:16
Aqueous Carlos
299313
asked Nov 16 '18 at 3:06
Durian JaykinDurian Jaykin
126
edited Nov 16 '18 at 3:16
Aqueous Carlos
299313
asked Nov 16 '18 at 3:06
Durian JaykinDurian Jaykin
126
edited Nov 16 '18 at 3:16
Aqueous Carlos
299313
edited Nov 16 '18 at 3:16
Aqueous Carlos
299313
edited Nov 16 '18 at 3:16
Aqueous Carlos
299313
299313
asked Nov 16 '18 at 3:06
Durian JaykinDurian Jaykin
126
asked Nov 16 '18 at 3:06
Durian JaykinDurian Jaykin
126
asked Nov 16 '18 at 3:06
Durian JaykinDurian Jaykin
126
126
add a comment |
add a comment |
1 Answer
1
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oldest
votes
Look at what the error is telling you - it's a different url. In particular, it's throwing an error because it ends with %20. This means there was a space at the end of your input. I would recommend processing your input to avoid this, something like
new_url = url.strip()
answered Nov 16 '18 at 3:45
tigerninjamantigerninjaman
10511
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
active
oldest
votes
active
oldest
votes
Look at what the error is telling you - it's a different url. In particular, it's throwing an error because it ends with %20. This means there was a space at the end of your input. I would recommend processing your input to avoid this, something like
new_url = url.strip()
answered Nov 16 '18 at 3:45
tigerninjamantigerninjaman
10511
add a comment |
Look at what the error is telling you - it's a different url. In particular, it's throwing an error because it ends with %20. This means there was a space at the end of your input. I would recommend processing your input to avoid this, something like
new_url = url.strip()
answered Nov 16 '18 at 3:45
tigerninjamantigerninjaman
10511
add a comment |
Look at what the error is telling you - it's a different url. In particular, it's throwing an error because it ends with %20. This means there was a space at the end of your input. I would recommend processing your input to avoid this, something like
new_url = url.strip()
answered Nov 16 '18 at 3:45
tigerninjamantigerninjaman
10511
Look at what the error is telling you - it's a different url. In particular, it's throwing an error because it ends with %20. This means there was a space at the end of your input. I would recommend processing your input to avoid this, something like
new_url = url.strip()
answered Nov 16 '18 at 3:45
tigerninjamantigerninjaman
10511
answered Nov 16 '18 at 3:45
tigerninjamantigerninjaman
10511
answered Nov 16 '18 at 3:45
tigerninjamantigerninjaman
10511
answered Nov 16 '18 at 3:45
tigerninjamantigerninjaman
10511
10511
add a comment |
add a comment |
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