如何在此处提取图像的链接？答案

【问题标题】：How can I extract the link to the image here?如何在此处提取图像的链接？
【发布时间】：2021-10-09 05:22:27
【问题描述】：

我正在使用 BS4 抓取网页中的文章标题、作者姓名以及指向其封面图片的链接。

这是我为特定文章的图片找到的 HTML 标记：

<img src="https://cdn.theatlantic.com/thumbor/lYaNkJfiPr8_Jp5Pj5ikj94GJgQ=/183x0:2000x1125/210x130/media/img/2021/06/WEL_Tiffany_KodakOpenerAlt-1/original.jpg" alt="Photo collage on a yellow background" loading="lazy" class="Image_root__J8Wlz Image_lazy__1w_jB Image_loaded__3uNg2 LandingRiver_image__1ZCUb" width="210" height="130" srcset="https://cdn.theatlantic.com/thumbor/lYaNkJfiPr8_Jp5Pj5ikj94GJgQ=/183x0:2000x1125/210x130/media/img/2021/06/WEL_Tiffany_KodakOpenerAlt-1/original.jpg, https://cdn.theatlantic.com/thumbor/NL-ZZp6sHiX_1WX2bn9f9o1WWjk=/183x0:2000x1125/420x260/media/img/2021/06/WEL_Tiffany_KodakOpenerAlt-1/original.jpg 2x">

这是我用来获取图片 URL 的代码：

cover_image = card.find('img')['src']

但是，而不是

"https://cdn.theatlantic.com/thumbor/lYaNkJfiPr8_Jp5Pj5ikj94GJgQ=/183x0:2000x1125/210x130/media/img/2021/06/WEL_Tiffany_KodakOpenerAlt-1/original.jpg"

它不断返回

data:image/svg+xml,%3Csvg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 210 130' /%3E

我也尝试过获取标记的“srcset”部分，但它甚至没有出现在代码中。

在这种情况下我该怎么办？

【问题讨论】：

一个html中有多个img标签，find给你第一个匹配，试试class_="Image_root__J8Wlz Image_lazy__1w_jB Image_loaded__3uNg2 LandingRiver_image__1ZCUb"

标签： python html image web-scraping beautifulsoup

【解决方案1】：

尝试使用Epis95 建议的类名进行选择：

cover_image = card.find('img',class_ = "Image_root__J8Wlz Image_lazy__1w_jB Image_loaded__3uNg2 LandingRiver_image__1ZCUb")['src']

# and if you want the best image quality you could do

# firstly get the srcset
cover_image = card.find('img',class_ = "Image_root__J8Wlz Image_lazy__1w_jB Image_loaded__3uNg2 LandingRiver_image__1ZCUb")['srcset']

# as srcset is a string but the links are seperated using ',' you could split it on the same
cover_image = cover_image.split(',')

# after splitting the best image quality lik is the last one so get the last element of the new list that you have after splitting
cover_image = cover_image[-1]

# you now have a string with 2x as the unwanted part at the last and you could remove that using the slice operation
cover_image = cover_image[:-2]    # slice it upto the third last characterto chop off '2x'

# now you have a string with leading and trailing spaces so strip it to remove any leading or trailing spaces
cover_image = cover_image.strip()
print(cover_image)    # this is the best quality image you can get

输出：

https://cdn.theatlantic.com/thumbor/NL-ZZp6sHiX_1WX2bn9f9o1WWjk=/183x0:2000x1125/420x260/media/img/2021/06/WEL_Tiffany_KodakOpenerAlt-1/original.jpg

【讨论】：