【发布时间】:2016-10-13 16:40:13
【问题描述】:
我想为所有参展商抓取此页面:
https://greenbuildexpo.com/Attendee/Expohall/Exhibitors
但是scrapy不加载内容,我现在正在做的是使用selenium来加载页面并用scrapy搜索链接:
url = 'https://greenbuildexpo.com/Attendee/Expohall/Exhibitors'
driver_1 = webdriver.Firefox()
driver_1.get(url)
content = driver_1.page_source
response = TextResponse(url='',body=content,encoding='utf-8')
print len(set(response.xpath('//*[contains(@href,"Attendee/")]//@href').extract()))
当按下“下一步”按钮时,该网站似乎没有发出任何新请求,所以我希望同时获得所有链接,但我只获得了 43 个带有该代码的链接。他们应该在 500 左右。
现在我正在尝试通过按“下一步”按钮来抓取页面:
for i in range(10):
xpath = '//*[@id="pagingNormalView"]/ul/li[15]'
driver_1.find_element_by_xpath(xpath).click()
但我得到了一个错误:
File "/usr/local/lib/python2.7/dist-packages/selenium/webdriver/remote/errorhandler.py", line 192, in check_response
raise exception_class(message, screen, stacktrace)
selenium.common.exceptions.NoSuchElementException: Message: Unable to locate element: {"method":"xpath","selector":"//*[@id=\"pagingNormalView\"]/ul/li[15]"}
Stacktrace:
【问题讨论】:
标签: python selenium xpath scrapy