ERROR requests return Response [503] If LINK link die or LINK live.I use bs4 python
i want crawl data from amazon,requests respone error 503,404,200 then i will check link die or live. But current link die or link live or link not found will respone 503.i dont know how check link ?.if respone 200 then will clear for software of me. THANKS YOUR HELP ME!!!!
link = "https://www.amazon.com/dp/B07K896272"enter code here
browser = webdriver.Firefox(executable_path=r'D:PythonToolAmzToolgeckodriver.exe')
browser.get(link)
res = requests.get(str(link).strip())
print(str(res))
python-3.x beautifulsoup python-requests amazon
asked Nov 16 at 2:27
Bao Quyet Nguyen
152
add a comment |
i want crawl data from amazon,requests respone error 503,404,200 then i will check link die or live. But current link die or link live or link not found will respone 503.i dont know how check link ?.if respone 200 then will clear for software of me. THANKS YOUR HELP ME!!!!
link = "https://www.amazon.com/dp/B07K896272"enter code here
browser = webdriver.Firefox(executable_path=r'D:PythonToolAmzToolgeckodriver.exe')
browser.get(link)
res = requests.get(str(link).strip())
print(str(res))
python-3.x beautifulsoup python-requests amazon
asked Nov 16 at 2:27
Bao Quyet Nguyen
152
add a comment |
i want crawl data from amazon,requests respone error 503,404,200 then i will check link die or live. But current link die or link live or link not found will respone 503.i dont know how check link ?.if respone 200 then will clear for software of me. THANKS YOUR HELP ME!!!!
link = "https://www.amazon.com/dp/B07K896272"enter code here
browser = webdriver.Firefox(executable_path=r'D:PythonToolAmzToolgeckodriver.exe')
browser.get(link)
res = requests.get(str(link).strip())
print(str(res))
python-3.x beautifulsoup python-requests amazon
asked Nov 16 at 2:27
Bao Quyet Nguyen
152
i want crawl data from amazon,requests respone error 503,404,200 then i will check link die or live. But current link die or link live or link not found will respone 503.i dont know how check link ?.if respone 200 then will clear for software of me. THANKS YOUR HELP ME!!!!
link = "https://www.amazon.com/dp/B07K896272"enter code here
browser = webdriver.Firefox(executable_path=r'D:PythonToolAmzToolgeckodriver.exe')
browser.get(link)
res = requests.get(str(link).strip())
print(str(res))
python-3.x beautifulsoup python-requests amazon
python-3.x beautifulsoup python-requests amazon
asked Nov 16 at 2:27
Bao Quyet Nguyen
152
asked Nov 16 at 2:27
Bao Quyet Nguyen
152
asked Nov 16 at 2:27
Bao Quyet Nguyen
152
asked Nov 16 at 2:27
Bao Quyet Nguyen
152
asked Nov 16 at 2:27
Bao Quyet Nguyen
152
152
add a comment |
add a comment |
1 Answer
1
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oldest
votes
If you hardcoded the string into variable link, you dont need to cast it to type str.
requests.get(link) // is good enough.
Also, if you want to print the response content,
print(res.text)
Don't really understand the question, but
you can check the response status by:
res = requests.get(link)
if res.status_code:
#Bad Code - 400s/500s
else:
#All good
Also, somes sites doesn't allow requests. You can try to be more "human" by adding in headers with user-agent and also use session. Session will keep the cookies. (kinda making it stateful)
session = requests.session()
session.headers['User-Agent'] = "YOUR USER AGENT HERE"
session.get("https://www.amazon.com/")
res = session.get(link)
print(res.text)
Some sites requires javascript to load the page. If that's the case, you will want to use selenium. using requests will not load the javascript page.
Or if u want to load the page with javascript first, and want to use requests:
session = requests.session()
session.headers['User-Agent'] = "YOUR USER AGENT HERE"
browser = webdriver.Firefox(executable_path=r'D:PythonToolAmzToolgeckodriver.exe')
browser.get(link)
for cookie in driver.get_cookies():
c = {cookie['name']: cookie['value']}
session.cookies.update(c)
browser.close()
res = session.get(link)
print(res.text)
answered Nov 20 at 1:13
NgoCuong
33917
thanks you many :v i understand and try solution of you
– Bao Quyet Nguyen
Nov 26 at 14:20
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
If you hardcoded the string into variable link, you dont need to cast it to type str.
requests.get(link) // is good enough.
Also, if you want to print the response content,
print(res.text)
Don't really understand the question, but
you can check the response status by:
res = requests.get(link)
if res.status_code:
#Bad Code - 400s/500s
else:
#All good
Also, somes sites doesn't allow requests. You can try to be more "human" by adding in headers with user-agent and also use session. Session will keep the cookies. (kinda making it stateful)
session = requests.session()
session.headers['User-Agent'] = "YOUR USER AGENT HERE"
session.get("https://www.amazon.com/")
res = session.get(link)
print(res.text)
Some sites requires javascript to load the page. If that's the case, you will want to use selenium. using requests will not load the javascript page.
Or if u want to load the page with javascript first, and want to use requests:
session = requests.session()
session.headers['User-Agent'] = "YOUR USER AGENT HERE"
browser = webdriver.Firefox(executable_path=r'D:PythonToolAmzToolgeckodriver.exe')
browser.get(link)
for cookie in driver.get_cookies():
c = {cookie['name']: cookie['value']}
session.cookies.update(c)
browser.close()
res = session.get(link)
print(res.text)
answered Nov 20 at 1:13
NgoCuong
33917
thanks you many :v i understand and try solution of you
– Bao Quyet Nguyen
Nov 26 at 14:20
add a comment |
If you hardcoded the string into variable link, you dont need to cast it to type str.
requests.get(link) // is good enough.
Also, if you want to print the response content,
print(res.text)
Don't really understand the question, but
you can check the response status by:
res = requests.get(link)
if res.status_code:
#Bad Code - 400s/500s
else:
#All good
Also, somes sites doesn't allow requests. You can try to be more "human" by adding in headers with user-agent and also use session. Session will keep the cookies. (kinda making it stateful)
session = requests.session()
session.headers['User-Agent'] = "YOUR USER AGENT HERE"
session.get("https://www.amazon.com/")
res = session.get(link)
print(res.text)
Some sites requires javascript to load the page. If that's the case, you will want to use selenium. using requests will not load the javascript page.
Or if u want to load the page with javascript first, and want to use requests:
session = requests.session()
session.headers['User-Agent'] = "YOUR USER AGENT HERE"
browser = webdriver.Firefox(executable_path=r'D:PythonToolAmzToolgeckodriver.exe')
browser.get(link)
for cookie in driver.get_cookies():
c = {cookie['name']: cookie['value']}
session.cookies.update(c)
browser.close()
res = session.get(link)
print(res.text)
answered Nov 20 at 1:13
NgoCuong
33917
thanks you many :v i understand and try solution of you
– Bao Quyet Nguyen
Nov 26 at 14:20
add a comment |
If you hardcoded the string into variable link, you dont need to cast it to type str.
requests.get(link) // is good enough.
Also, if you want to print the response content,
print(res.text)
Don't really understand the question, but
you can check the response status by:
res = requests.get(link)
if res.status_code:
#Bad Code - 400s/500s
else:
#All good
Also, somes sites doesn't allow requests. You can try to be more "human" by adding in headers with user-agent and also use session. Session will keep the cookies. (kinda making it stateful)
session = requests.session()
session.headers['User-Agent'] = "YOUR USER AGENT HERE"
session.get("https://www.amazon.com/")
res = session.get(link)
print(res.text)
Some sites requires javascript to load the page. If that's the case, you will want to use selenium. using requests will not load the javascript page.
Or if u want to load the page with javascript first, and want to use requests:
session = requests.session()
session.headers['User-Agent'] = "YOUR USER AGENT HERE"
browser = webdriver.Firefox(executable_path=r'D:PythonToolAmzToolgeckodriver.exe')
browser.get(link)
for cookie in driver.get_cookies():
c = {cookie['name']: cookie['value']}
session.cookies.update(c)
browser.close()
res = session.get(link)
print(res.text)
answered Nov 20 at 1:13
NgoCuong
33917
If you hardcoded the string into variable link, you dont need to cast it to type str.
requests.get(link) // is good enough.
Also, if you want to print the response content,
print(res.text)
Don't really understand the question, but
you can check the response status by:
res = requests.get(link)
if res.status_code:
#Bad Code - 400s/500s
else:
#All good
Also, somes sites doesn't allow requests. You can try to be more "human" by adding in headers with user-agent and also use session. Session will keep the cookies. (kinda making it stateful)
session = requests.session()
session.headers['User-Agent'] = "YOUR USER AGENT HERE"
session.get("https://www.amazon.com/")
res = session.get(link)
print(res.text)
Some sites requires javascript to load the page. If that's the case, you will want to use selenium. using requests will not load the javascript page.
Or if u want to load the page with javascript first, and want to use requests:
session = requests.session()
session.headers['User-Agent'] = "YOUR USER AGENT HERE"
browser = webdriver.Firefox(executable_path=r'D:PythonToolAmzToolgeckodriver.exe')
browser.get(link)
for cookie in driver.get_cookies():
c = {cookie['name']: cookie['value']}
session.cookies.update(c)
browser.close()
res = session.get(link)
print(res.text)
answered Nov 20 at 1:13
NgoCuong
33917
edited Nov 20 at 7:36
answered Nov 20 at 1:13
NgoCuong
33917
answered Nov 20 at 1:13
NgoCuong
33917
answered Nov 20 at 1:13
NgoCuong
33917
33917
thanks you many :v i understand and try solution of you
– Bao Quyet Nguyen
Nov 26 at 14:20
add a comment |
thanks you many :v i understand and try solution of you
– Bao Quyet Nguyen
Nov 26 at 14:20
thanks you many :v i understand and try solution of you
– Bao Quyet Nguyen
Nov 26 at 14:20
thanks you many :v i understand and try solution of you
– Bao Quyet Nguyen
Nov 26 at 14:20
add a comment |
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