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Unable to fetch some links using list comprehension within scrapy


Generator Expressions vs. List ComprehensionList comprehension vs mapCreate a dictionary with list comprehension in Pythonlist comprehension vs. lambda + filterif/else in Python's list comprehension?if else in a list comprehensionScrapy - parse a page to extract items - then follow and store item url contentsScrapy: Get data on page and following linkHow to keep track of a request in scrapyPython Scrapy and yielding













0















I've written a script in python using scrapy to get the links from response after making a post request to a certain url. The links are perfectly coming through when I try with the following script.



Working one:



import scrapy
from scrapy.crawler import CrawlerProcess

class AftnetSpider(scrapy.Spider):
name = "aftnet"
base_url = "http://www.aftnet.be/MyAFT/Clubs/SearchClubs"

def start_requests(self):
yield scrapy.FormRequest(self.base_url,callback=self.parse,formdata='regions':'1,3,4,6')

def parse(self,response):
for items in response.css("dl.club-item"):
for item in items.css("dd a[data-toggle='popover']::attr('data-url')").getall():
yield "result_url":response.urljoin(item)

if __name__ == "__main__":
c = CrawlerProcess(
'USER_AGENT': 'Mozilla/5.0',

)
c.crawl(AftnetSpider)
c.start()


However, my intention is to achieve the same using list comprehension but I'm getting some error.



Using list comprehension:



def parse(self,response):
return [response.urljoin(item) for items in response.css("dl.club-item") for item in items.css("dd a[data-toggle='popover']::attr('data-url')").getall()]


I get the following error:



2019-03-08 12:45:44 [scrapy.core.scraper] ERROR: Spider must return Request, BaseItem, dict or None, got 'str' in <POST http://www.aftnet.be/MyAFT/Clubs/SearchClubs>



How can I get some links using list comprehension within scrapy?











share|improve this question


























    0















    I've written a script in python using scrapy to get the links from response after making a post request to a certain url. The links are perfectly coming through when I try with the following script.



    Working one:



    import scrapy
    from scrapy.crawler import CrawlerProcess

    class AftnetSpider(scrapy.Spider):
    name = "aftnet"
    base_url = "http://www.aftnet.be/MyAFT/Clubs/SearchClubs"

    def start_requests(self):
    yield scrapy.FormRequest(self.base_url,callback=self.parse,formdata='regions':'1,3,4,6')

    def parse(self,response):
    for items in response.css("dl.club-item"):
    for item in items.css("dd a[data-toggle='popover']::attr('data-url')").getall():
    yield "result_url":response.urljoin(item)

    if __name__ == "__main__":
    c = CrawlerProcess(
    'USER_AGENT': 'Mozilla/5.0',

    )
    c.crawl(AftnetSpider)
    c.start()


    However, my intention is to achieve the same using list comprehension but I'm getting some error.



    Using list comprehension:



    def parse(self,response):
    return [response.urljoin(item) for items in response.css("dl.club-item") for item in items.css("dd a[data-toggle='popover']::attr('data-url')").getall()]


    I get the following error:



    2019-03-08 12:45:44 [scrapy.core.scraper] ERROR: Spider must return Request, BaseItem, dict or None, got 'str' in <POST http://www.aftnet.be/MyAFT/Clubs/SearchClubs>



    How can I get some links using list comprehension within scrapy?











    share|improve this question
























      0












      0








      0








      I've written a script in python using scrapy to get the links from response after making a post request to a certain url. The links are perfectly coming through when I try with the following script.



      Working one:



      import scrapy
      from scrapy.crawler import CrawlerProcess

      class AftnetSpider(scrapy.Spider):
      name = "aftnet"
      base_url = "http://www.aftnet.be/MyAFT/Clubs/SearchClubs"

      def start_requests(self):
      yield scrapy.FormRequest(self.base_url,callback=self.parse,formdata='regions':'1,3,4,6')

      def parse(self,response):
      for items in response.css("dl.club-item"):
      for item in items.css("dd a[data-toggle='popover']::attr('data-url')").getall():
      yield "result_url":response.urljoin(item)

      if __name__ == "__main__":
      c = CrawlerProcess(
      'USER_AGENT': 'Mozilla/5.0',

      )
      c.crawl(AftnetSpider)
      c.start()


      However, my intention is to achieve the same using list comprehension but I'm getting some error.



      Using list comprehension:



      def parse(self,response):
      return [response.urljoin(item) for items in response.css("dl.club-item") for item in items.css("dd a[data-toggle='popover']::attr('data-url')").getall()]


      I get the following error:



      2019-03-08 12:45:44 [scrapy.core.scraper] ERROR: Spider must return Request, BaseItem, dict or None, got 'str' in <POST http://www.aftnet.be/MyAFT/Clubs/SearchClubs>



      How can I get some links using list comprehension within scrapy?











      share|improve this question














      I've written a script in python using scrapy to get the links from response after making a post request to a certain url. The links are perfectly coming through when I try with the following script.



      Working one:



      import scrapy
      from scrapy.crawler import CrawlerProcess

      class AftnetSpider(scrapy.Spider):
      name = "aftnet"
      base_url = "http://www.aftnet.be/MyAFT/Clubs/SearchClubs"

      def start_requests(self):
      yield scrapy.FormRequest(self.base_url,callback=self.parse,formdata='regions':'1,3,4,6')

      def parse(self,response):
      for items in response.css("dl.club-item"):
      for item in items.css("dd a[data-toggle='popover']::attr('data-url')").getall():
      yield "result_url":response.urljoin(item)

      if __name__ == "__main__":
      c = CrawlerProcess(
      'USER_AGENT': 'Mozilla/5.0',

      )
      c.crawl(AftnetSpider)
      c.start()


      However, my intention is to achieve the same using list comprehension but I'm getting some error.



      Using list comprehension:



      def parse(self,response):
      return [response.urljoin(item) for items in response.css("dl.club-item") for item in items.css("dd a[data-toggle='popover']::attr('data-url')").getall()]


      I get the following error:



      2019-03-08 12:45:44 [scrapy.core.scraper] ERROR: Spider must return Request, BaseItem, dict or None, got 'str' in <POST http://www.aftnet.be/MyAFT/Clubs/SearchClubs>



      How can I get some links using list comprehension within scrapy?








      python python-3.x web-scraping scrapy






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Mar 8 at 7:03









      MITHUMITHU

      239217




      239217






















          1 Answer
          1






          active

          oldest

          votes


















          1














          Your generator with a loop is returning a single dict on every call:



          yield "result_url":response.urljoin(item)


          But your list comprehension is returning a list of strings. I don't know why you want a list comprehension here: your generator is much easier to understand (as shown by the fact that you have got it to work and are having trouble with the list comprehension) but if you insist on doing it, what you need is a list of dicts not strings, something like



          return ["result_url":response.urljoin(item) for items in response.css("dl.club-item") for item in items.css("dd a[data-toggle='popover']::attr('data-url')").getall()]


          But please don't do that. Remember that readability counts. Your generator is readable, your one-liner isn't.






          share|improve this answer






















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            1 Answer
            1






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

            votes









            1














            Your generator with a loop is returning a single dict on every call:



            yield "result_url":response.urljoin(item)


            But your list comprehension is returning a list of strings. I don't know why you want a list comprehension here: your generator is much easier to understand (as shown by the fact that you have got it to work and are having trouble with the list comprehension) but if you insist on doing it, what you need is a list of dicts not strings, something like



            return ["result_url":response.urljoin(item) for items in response.css("dl.club-item") for item in items.css("dd a[data-toggle='popover']::attr('data-url')").getall()]


            But please don't do that. Remember that readability counts. Your generator is readable, your one-liner isn't.






            share|improve this answer



























              1














              Your generator with a loop is returning a single dict on every call:



              yield "result_url":response.urljoin(item)


              But your list comprehension is returning a list of strings. I don't know why you want a list comprehension here: your generator is much easier to understand (as shown by the fact that you have got it to work and are having trouble with the list comprehension) but if you insist on doing it, what you need is a list of dicts not strings, something like



              return ["result_url":response.urljoin(item) for items in response.css("dl.club-item") for item in items.css("dd a[data-toggle='popover']::attr('data-url')").getall()]


              But please don't do that. Remember that readability counts. Your generator is readable, your one-liner isn't.






              share|improve this answer

























                1












                1








                1







                Your generator with a loop is returning a single dict on every call:



                yield "result_url":response.urljoin(item)


                But your list comprehension is returning a list of strings. I don't know why you want a list comprehension here: your generator is much easier to understand (as shown by the fact that you have got it to work and are having trouble with the list comprehension) but if you insist on doing it, what you need is a list of dicts not strings, something like



                return ["result_url":response.urljoin(item) for items in response.css("dl.club-item") for item in items.css("dd a[data-toggle='popover']::attr('data-url')").getall()]


                But please don't do that. Remember that readability counts. Your generator is readable, your one-liner isn't.






                share|improve this answer













                Your generator with a loop is returning a single dict on every call:



                yield "result_url":response.urljoin(item)


                But your list comprehension is returning a list of strings. I don't know why you want a list comprehension here: your generator is much easier to understand (as shown by the fact that you have got it to work and are having trouble with the list comprehension) but if you insist on doing it, what you need is a list of dicts not strings, something like



                return ["result_url":response.urljoin(item) for items in response.css("dl.club-item") for item in items.css("dd a[data-toggle='popover']::attr('data-url')").getall()]


                But please don't do that. Remember that readability counts. Your generator is readable, your one-liner isn't.







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Mar 8 at 8:31









                BoarGulesBoarGules

                8,43721228




                8,43721228





























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