标签:
lis = response.xpath("//ul/li") for li in lis: src = li.xpath("img/@src") # 如果xpath表达式是"//img/@src"会把整个页面的所有图片src提取出来 alt = li.xpath("img/@alt")
scrapy xpath 从response中获取li,然后再获取li中img的src
原文地址:http://www.cnblogs.com/bushe/p/4462279.html