我目前正在尝试从以下网站上的每家餐厅抓取 href 元素：
https://www.menulog.com.au/area/2173-moorebank?lat=-33.9477825&long=150.9190988&q=liverpool
相关的html可以在：
但是，当我在 scrapy shell 中使用以下代码时，它什么也不返回
response.css("div.c-listing>div>div")
我想知道为什么会这样/我可以做些什么来解决这个问题？
谢谢！
【问题讨论】：
标签： python web-scraping scrapy
使用这个css选择器'a.c-listing-item-link.u-clearfix'在scrapy shell中提取url链接。
>>> for url in response.css('a.c-listing-item-link.u-clearfix ::attr("href")').extract():
... print(response.urljoin(url))
...
https://www.menulog.com.au/restaurants-blazin-grillz-liverpool/menu
https://www.menulog.com.au/restaurants-phillies-liverpool/menu
https://www.menulog.com.au/restaurants-mcdonalds-liverpool-south/menu
https://www.menulog.com.au/restaurants-kfc-liverpool/menu
https://www.menulog.com.au/restaurants-omer-biryani-house-liverpool/menu
https://www.menulog.com.au/restaurants-classic-burger-liverpool/menu
https://www.menulog.com.au/restaurants-jasmin-1-liverpool/menu
https://www.menulog.com.au/restaurants-subway-liverpool/menu
https://www.menulog.com.au/restaurants-himalayas-indian-restaurant-liverpool/menu
https://www.menulog.com.au/restaurants-jasmins-liverpool/menu
https://www.menulog.com.au/restaurants-sharetea-liverpool/menu
https://www.menulog.com.au/restaurants-healthy-kitchen-liverpool-halal/menu
https://www.menulog.com.au/restaurants-dosa-hut-liverpool/menu
https://www.menulog.com.au/restaurants-the-kulcha-house-liverpool/menu
https://www.menulog.com.au/restaurants-biang-biang-noodle-shop-liverpool/menu
https://www.menulog.com.au/restaurants-zambeekas-liverpool/menu
https://www.menulog.com.au/restaurants-mina-bakery-liverpool/menu
https://www.menulog.com.au/restaurants-crossroads-hotel-liverpool/menu
https://www.menulog.com.au/restaurants-nutrition-station-liverpool/menu
https://www.menulog.com.au/restaurants-mizuki-sushi-liverpool/menu
【讨论】：