Python 将lxml转换为scrapy xxs选择器
如何将这个纯python lxml转换为scrapy内置的xxs选择器?这一个工作,但我想把它转换为刮xxs选择器Python 将lxml转换为scrapy xxs选择器,python,xml,screen-scraping,scrapy,lxml,Python,Xml,Screen Scraping,Scrapy,Lxml,如何将这个纯python lxml转换为scrapy内置的xxs选择器?这一个工作,但我想把它转换为刮xxs选择器 def parse_device_list(self, response): self.log("\n\n\n List of devices \n\n\n") self.log('Hi, this is the parse_device_list page! %s' % response.url) root = lxml.etree.fromstr
def parse_device_list(self, response):
self.log("\n\n\n List of devices \n\n\n")
self.log('Hi, this is the parse_device_list page! %s' % response.url)
root = lxml.etree.fromstring(response.body)
for row in root.xpath('//row'):
allcells = row.xpath('./cell')
# first cell contain the link to follow
detail_page_link = allcells[0].get("href")
yield Request(urlparse.urljoin(response.url, detail_page_link ), callback=self.parse_page)
试一试:
def parse_page(self, response):
xxs = XmlXPathSelector(response)
for row in xxs.select('//row'):
detail_page_link = row.select('.//cell[1]/@href')[0].extract()
yield Request(urlparse.urljoin(response.url, detail_page_link), callback=self.parse_page)
这似乎是可行的,但我如何让它按顺序迭代呢?出于某种原因,它迭代了列A,但顺序不正确。当列A中的一行为空时,它将获取列B链接。我可以让我的方法只获取A列,如果A列为null,则跳过它并转到A列的下一行。如果A列为空,则不获取B列。