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POUsing Scrapy to parse site, follow Next Page, write as XML
 

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    <p>My script works wonderfully when I comment one piece of code: <strong>return items</strong>.</p> <p>Here is my code, changing to <a href="http://example.com" rel="nofollow">http://example.com</a> since it appears that is what other people to possibly to preserve the 'scraping' legality issues.</p> <pre><code>class Vfood(CrawlSpider): name = "example.com" allowed_domains = [ "example.com" ] start_urls = [ "http://www.example.com/TV_Shows/Show/Episodes" ] rules = ( Rule(SgmlLinkExtractor(allow=('example\.com', 'page='), restrict_xpaths = '//div[@class="paginator"]/ span[@id="next"]'), callback='parse'), ) def parse(self, response): hxs = HtmlXPathSelector(response) items = [] countries = hxs.select('//div[@class="index-content"]') tmpNextPage = hxs.select('//div[@class="paginator"]/span[@id="next"]/a/@href').extract() for country in countries: item = FoodItem() countryName = country.select('.//h3/text()').extract() item['country'] = countryName print "Country Name: ", countryName shows = country.select('.//div[@class="content1"]') for show in shows.select('.//div'): showLink = (show.select('.//h4/a/@href').extract()).pop() showLocation = show.select('.//h4/a/text()').extract() showText = show.select('.//p/text()').extract() item['showURL'] = "http://www.travelchannel.com"+str(showLink) item['showcity'] = showLocation item['showtext'] = showText item['showtext'] = showText print "\t", showLink print "\t", showLocation print "\t", showText print "\n" items.append(item) **#return items** for NextPageLink in tmpNextPage: m = re.search("Location", NextPageLink) if m: NextPage = NextPageLink print "Next Page: ", NextPage yield Request("http://www.example.com/"+NextPage, callback = self.parse) else: NextPage = 'None' SPIDER = food() </code></pre> <p>If I UNCOMMENT the #return items, I get the following error: </p> <pre><code>yield Request("http://www.example.com/"+NextPage, callback = self.parse) SyntaxError: 'return' with argument inside generator </code></pre> <p>By leaving the comment there, I am unable to collect the data in XML format, but by the result of the print statements, I do see everything that I am supposed to on the screen.</p> <p>my command for getting xml out:</p> <pre><code>scrapy crawl example.com --set FEED_URI=food.xml --set FEED_FORMAT=xml </code></pre> <p>I get the XML file creation when I UNCOMMENT the <strong><em>return items</em></strong> line above, but the script stops and won't follow the links.</p>
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