小白求助：请问怎么爬取img标签下的src地址？

https://www.gooood.cn/sl_release-apartment-by-pascali-semerdjian-arquitetos.htm

img这个标签好像没有独特的元素去定位，比如每张图片的img标签下的class属性值都不同，只会用find_all('a', class_='colorbox_gallery')到，要取出里面img里的src就没招了，求指导

然后，下载图片的过程中，有时候会遇到UnicodeEncodeError: 'gbk' codec can't encode character '\u0131' in position 7: illegal multibyte sequence
图片下载就中断了，有解决的办法吗？

import requests
from lxml import etree
url = "https://www.gooood.cn/sl_release-apartment-by-pascali-semerdjian-arquitetos.htm"
r = requests.get(url)
print(r.content.decode())
html = etree.HTML(r.content.decode())
imgs = html.xpath("//img//@src")
n = 1
for i in imgs:
print(i)
response = requests.get(i)
img = response.content
with open("./imags/{}.jpg".format(n),"wb") as f: #需要在当前目录下建立imags文件夹
f.write(img)
n += 1

#亲测运行正常

UnicodeEncodeError:这个是网页的编码和你指定的不同，这个网站我打开看了下

<meta charset="utf-8">

是utf8的，你用utf8来decode，而不是 gbk

至于没有办法定位，你可以f12以后找到，复制生成的xpath，照着写

import urllib.request
from bs4 import BeautifulSoup

headers = {"User-Agent":"Mozilla/5.0 (Windows NT 6.3; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/59.0.3071.86 Safari/537.36"}
url="URL地址"

req=urllib.request.Request(url, headers= headers)

data=urllib.request.urlopen(req).read()

soup = BeautifulSoup(data, "html.parser")

links = soup.find_all('a', class_='colorbox_gallery')

imgs=[]

for link in links:
    imgs.append(link.find('img'))

for img in imgs:
    print(img)

假设find('a')储存在dat里

img = dat.find('img')
href = img.get('src')

href  即为链接

另外,图片是二进制的,用'wb'形式打开,写入要用  f.write(respons.content)