![](/img/trans.png)
[英]Get url of link using Python web scraping; requests, requests_html, selenium
[英]Parse an img src url using requests_html on python
我正在嘗試使用 HTML ZA2F21ED4F8EBC2CBB14 參考 Python 獲取 img src URL
我的代碼現在看起來像這樣,這些是我一直在使用的庫:
from requests_html import HTMLSession
import concurrent.futures
import sys
import sqlite3
data = {'url': url}
resp = sess.get(url)
artwork = resp.html.find('.single-album-tombstone__art img')
if artwork:
data['artwork'] = artwork[0].text
這不起作用,它只返回一個“EMPTY”字符串。 如何獲取圖像 URL? 我試過使用.attrs['src']
但也沒有成功。
你可以像這樣從img
標簽解析src
屬性
from requests_html import HTMLSession
from bs4 import BeautifulSoup
url = "https://www.google.com/search?q=james+bond&safe=active&tbm=isch"
with HTMLSession() as session:
response = session.get(url)
response.html.render()
content = response.html.html
soup = BeautifulSoup(content, "html.parser")
images = soup.find_all("img")
for img in images:
src = img.get("src")
if src:
print(src)
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.