使用scrapy获取下一页
我有兴趣从此页面获取亚特兰大的承包商数据: 因此,我可以打开类别的链接 “增加和改造”使用scrapy获取下一页,scrapy,web-crawler,Scrapy,Web Crawler,我有兴趣从此页面获取亚特兰大的承包商数据: 因此,我可以打开类别的链接 “增加和改造” “建筑师与工程师” “喷泉和池塘” …… ..… ..... 但我只能打开第一页: 我正试图通过“下一步”按钮的链接打开“获取下一步”: next_page_url = response.xpath('/html/body/div[1]/center/table/tr[8]/td[2]/a/@href').extract_first() absolute_next_page_url =
“建筑师与工程师”
“喷泉和池塘”
……
..…
..... 但我只能打开第一页: 我正试图通过“下一步”按钮的链接打开“获取下一步”:
next_page_url = response.xpath('/html/body/div[1]/center/table/tr[8]/td[2]/a/@href').extract_first()
absolute_next_page_url = response.urljoin(next_page_url)
request = scrapy.Request(absolute_next_page_url)
yield request
但这没什么区别
这是我的蜘蛛的代码:
import scrapy
class Spider_1800(scrapy.Spider):
name = '1800contractor'
allowed_domains = ['1800contractor.com']
start_urls = (
'http://www.1800contractor.com/d.Atlanta.GA.html?link_id=3658',
)
def parse(self, response):
urls = response.xpath('/html/body/center/table/tr/td[2]/table/tr[6]/td/table/tr[2]/td/b/a/@href').extract()
for url in urls:
absolute_url = response.urljoin(url)
request = scrapy.Request(
absolute_url, callback=self.parse_contractors)
yield request
# process next page
next_page_url = response.xpath('/html/body/div[1]/center/table/tr[8]/td[2]/a/@href').extract_first()
absolute_next_page_url = response.urljoin(next_page_url)
request = scrapy.Request(absolute_next_page_url)
yield request
def parse_contractors(self, response):
name = response.xpath(
'/html/body/div[1]/center/table/tr[5]/td/table/tr[1]/td/b/a/@href').extract()
contrator = {
'name': name,
'url': response.url}
yield contrator
如果没有为正确的请求分页,
parse
将处理使用start\u url
中的URL生成的请求,这意味着您需要首先输入每个类别
点击开始url后,为承包商选择url的xpath不起作用。下一页出现在承包商页面上,因此在承包商url后调用。这对你有用
def parse(self, response):
urls = response.xpath('//table//*[@class="hiCatNaked"]').extract()
for url in urls:
absolute_url = response.urljoin(url)
request = scrapy.Request(
absolute_url, callback=self.parse_contractors)
yield request
def parse_contractors(self, response):
name=response.xpath('/html/body/div[1]/center/table/tr[5]/td/table/tr[1]/td/b/a/@href').extract()
contrator = {
'name': name,
'url': response.url}
yield contrator
next_page_url = response.xpath('//a[b[contains(.,'Next')]]/@href').extract_first()
if next_page_url:
absolute_next_page_url = response.urljoin(next_page_url)
yield scrapy.Request(absolute_next_page_url, callback=self.parse_contractors)
我仍然得到相同的结果请检查更新的答案,看起来您在回调方法上处理了错误的请求
def parse(self, response):
urls = response.xpath('//table//*[@class="hiCatNaked"]').extract()
for url in urls:
absolute_url = response.urljoin(url)
request = scrapy.Request(
absolute_url, callback=self.parse_contractors)
yield request
def parse_contractors(self, response):
name=response.xpath('/html/body/div[1]/center/table/tr[5]/td/table/tr[1]/td/b/a/@href').extract()
contrator = {
'name': name,
'url': response.url}
yield contrator
next_page_url = response.xpath('//a[b[contains(.,'Next')]]/@href').extract_first()
if next_page_url:
absolute_next_page_url = response.urljoin(next_page_url)
yield scrapy.Request(absolute_next_page_url, callback=self.parse_contractors)