我必须从这个网址一一点击每个搜索结果:
我首先从显示的文本中提取结果总数,以便我可以设置迭代的上限
upperlimit=driver.find_element_by_id("total_results")
number = int(upperlimit.text.split(' ')[0])
循环被定义为 对于范围内的 i(1,数字):
但是,在浏览完第一页上的前 10 个结果后,列表索引超出范围(可能是因为没有更多链接可供单击)。我需要单击“下一步”才能获取接下来的 10 个结果,依此类推,直到完成所有搜索结果。我怎样才能做到这一点?
如有任何帮助,我们将不胜感激!
最佳答案
问题在于 id 为 total_results
的元素的值页面加载后发生变化,首先包含 117
,然后更改为 44
.
相反,这里有一个更强大的方法。它逐页处理,直到没有更多页面为止:
from selenium import webdriver
from selenium.common.exceptions import NoSuchElementException
driver = webdriver.Firefox()
url = 'http://www.nice.org.uk/Search.do?searchText=bevacizumab&newsearch=true#/search/?searchText=bevacizumab&mode=&staticTitle=false&SEARCHTYPE_all2=true&SEARCHTYPE_all1=&SEARCHTYPE=GUIDANCE&TOPICLVL0_all2=true&TOPICLVL0_all1=&HIDEFILTER=TOPICLVL1&HIDEFILTER=TOPICLVL2&TREATMENTS_all2=true&TREATMENTS_all1=&GUIDANCETYPE_all2=true&GUIDANCETYPE_all1=&STATUS_all2=true&STATUS_all1=&HIDEFILTER=EGAPREFERENCE&HIDEFILTER=TOPICLVL3&DATEFILTER_ALL=ALL&DATEFILTER_PREV=ALL&custom_date_from=&custom_date_to=11-06-2014&PAGINATIONURL=%2FSearch.do%3FsearchText%40%40bevacizumab%26newsearch%40%40true%26page%40%40&SORTORDER=BESTMATCH'
driver.get(url)
page_number = 1
while True:
try:
link = driver.find_element_by_link_text(str(page_number))
except NoSuchElementException:
break
link.click()
print driver.current_url
page_number += 1
基本上,这里的想法是获取下一页链接,直到没有这样的链接( NoSuchElementException
将被抛出)。请注意,它适用于任意数量的页面和结果。
它打印:
http://www.nice.org.uk/Search.do?searchText=bevacizumab&newsearch=true&page=1
http://www.nice.org.uk/Search.do?searchText=bevacizumab&newsearch=true&page=2#showfilter
http://www.nice.org.uk/Search.do?searchText=bevacizumab&newsearch=true&page=3#showfilter
http://www.nice.org.uk/Search.do?searchText=bevacizumab&newsearch=true&page=4#showfilter
http://www.nice.org.uk/Search.do?searchText=bevacizumab&newsearch=true&page=5#showfilter
关于python - Selenium Python - 访问搜索结果的下一页,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/24166689/