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Scrapy Avoid External Links Missing Protocol
2019 Community Moderator ElectionCalling an external command in PythonScraping large number of sites with ScrapyCannot display HTML stringPyQt4 Scrapy ImplementationHow to get link from onclick event using Scrapy?Does Scrapy crawl ALL links with Rules?Scrapy login to vBulletin guidance neededScrapy Missing fields/data in output fileUsing Scrapy on a Google cache of a websiteTuning scrapy to avoid specific links and return url responses
I'm a total Scrapy n00b, and am encountering a problematic situation. Several pages on the site I'm scraping contain external links in the following format:
<a href="www.externalsite.com/somepage">www.externalsite.com/somepage.</a>
The problem is that because the protocol is missing from the link, Scrapy takes the completely reasonable action of harvesting the link and pre-pending the base domain onto it, resulting in a link like so:
https://www.basedomain.com/page1/www.externalsite.com/somepage
This is perfectly reasonable, as it's the same action a browser takes when you click the external link missing the protocol. The problem is that in Scrapy, this creates a spider trap following links like these, ad infinitum:
https://www.basedomain.com/page1/www.externalsite.com/somepage/www.externalsite.com/somepage
https://www.basedomain.com/page1/www.externalsite.com/somepage/www.externalsite.com/somepage/www.externalsite.com/somepage
Eventually the URL is so long the server returns 500 and the loop stops.
I know there must be a way to avoid this issue with the LinkExtractor, but I just don't know how to do it. And I would prefer to avoid hard-coding a case for this site, and find a solution that will work for this scenario regardless. Any information would be greatly appreciated.
python scrapy
add a comment |
I'm a total Scrapy n00b, and am encountering a problematic situation. Several pages on the site I'm scraping contain external links in the following format:
<a href="www.externalsite.com/somepage">www.externalsite.com/somepage.</a>
The problem is that because the protocol is missing from the link, Scrapy takes the completely reasonable action of harvesting the link and pre-pending the base domain onto it, resulting in a link like so:
https://www.basedomain.com/page1/www.externalsite.com/somepage
This is perfectly reasonable, as it's the same action a browser takes when you click the external link missing the protocol. The problem is that in Scrapy, this creates a spider trap following links like these, ad infinitum:
https://www.basedomain.com/page1/www.externalsite.com/somepage/www.externalsite.com/somepage
https://www.basedomain.com/page1/www.externalsite.com/somepage/www.externalsite.com/somepage/www.externalsite.com/somepage
Eventually the URL is so long the server returns 500 and the loop stops.
I know there must be a way to avoid this issue with the LinkExtractor, but I just don't know how to do it. And I would prefer to avoid hard-coding a case for this site, and find a solution that will work for this scenario regardless. Any information would be greatly appreciated.
python scrapy
add a comment |
I'm a total Scrapy n00b, and am encountering a problematic situation. Several pages on the site I'm scraping contain external links in the following format:
<a href="www.externalsite.com/somepage">www.externalsite.com/somepage.</a>
The problem is that because the protocol is missing from the link, Scrapy takes the completely reasonable action of harvesting the link and pre-pending the base domain onto it, resulting in a link like so:
https://www.basedomain.com/page1/www.externalsite.com/somepage
This is perfectly reasonable, as it's the same action a browser takes when you click the external link missing the protocol. The problem is that in Scrapy, this creates a spider trap following links like these, ad infinitum:
https://www.basedomain.com/page1/www.externalsite.com/somepage/www.externalsite.com/somepage
https://www.basedomain.com/page1/www.externalsite.com/somepage/www.externalsite.com/somepage/www.externalsite.com/somepage
Eventually the URL is so long the server returns 500 and the loop stops.
I know there must be a way to avoid this issue with the LinkExtractor, but I just don't know how to do it. And I would prefer to avoid hard-coding a case for this site, and find a solution that will work for this scenario regardless. Any information would be greatly appreciated.
python scrapy
I'm a total Scrapy n00b, and am encountering a problematic situation. Several pages on the site I'm scraping contain external links in the following format:
<a href="www.externalsite.com/somepage">www.externalsite.com/somepage.</a>
The problem is that because the protocol is missing from the link, Scrapy takes the completely reasonable action of harvesting the link and pre-pending the base domain onto it, resulting in a link like so:
https://www.basedomain.com/page1/www.externalsite.com/somepage
This is perfectly reasonable, as it's the same action a browser takes when you click the external link missing the protocol. The problem is that in Scrapy, this creates a spider trap following links like these, ad infinitum:
https://www.basedomain.com/page1/www.externalsite.com/somepage/www.externalsite.com/somepage
https://www.basedomain.com/page1/www.externalsite.com/somepage/www.externalsite.com/somepage/www.externalsite.com/somepage
Eventually the URL is so long the server returns 500 and the loop stops.
I know there must be a way to avoid this issue with the LinkExtractor, but I just don't know how to do it. And I would prefer to avoid hard-coding a case for this site, and find a solution that will work for this scenario regardless. Any information would be greatly appreciated.
python scrapy
python scrapy
asked Mar 6 at 19:50
LandonCLandonC
414620
414620
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