是否可以从 instagram url 获取图像 url?
Is possibile to get image url from instagram url?
我有一个 Instagram url:
https://www.instagram.com/p/B2EtjT9hgvG/
里面的图片有url:
如果我只有 Instagram url 是否可以使用 API 或其他方式获取图像 url?
TY
感谢@Chris Doyle 的帮助,我可以使用 selenium
:
为您提供一些帮助
from selenium import webdriver
driver = webdriver.Chrome('chromedriver.exe')
driver.get('https://www.instagram.com/p/B2EtjT9hgvG/')
metas = driver.find_elements_by_tag_name('meta')
for elem in metas:
if elem.get_attribute('property') == 'og:image':
print(elem.get_attribute('content'))
或相应地 requests
和 bs4
:
import requests
from bs4 import BeautifulSoup
result = requests.get("https://www.instagram.com/p/B2EtjT9hgvG/")
c = result.content
soup = BeautifulSoup(c)
metas = soup.find_all(attrs={"property": "og:image"})
print(metas[0].attrs['content'])
我有一个 Instagram url:
https://www.instagram.com/p/B2EtjT9hgvG/
里面的图片有url:
如果我只有 Instagram url 是否可以使用 API 或其他方式获取图像 url? TY
感谢@Chris Doyle 的帮助,我可以使用 selenium
:
from selenium import webdriver
driver = webdriver.Chrome('chromedriver.exe')
driver.get('https://www.instagram.com/p/B2EtjT9hgvG/')
metas = driver.find_elements_by_tag_name('meta')
for elem in metas:
if elem.get_attribute('property') == 'og:image':
print(elem.get_attribute('content'))
或相应地 requests
和 bs4
:
import requests
from bs4 import BeautifulSoup
result = requests.get("https://www.instagram.com/p/B2EtjT9hgvG/")
c = result.content
soup = BeautifulSoup(c)
metas = soup.find_all(attrs={"property": "og:image"})
print(metas[0].attrs['content'])